Normal Distribution (CIE AS Maths: Probability & Statistics 1)

Exam Questions

4 hours27 questions
1a
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1 mark

A continuous random variable can take any value within a given range.  Many naturally occurring continuous quantities can be modelled using the Normal Distribution, for example the height of human beings; the mass of new born puppies or the distribution of all A Level maths exam results.

Give a different example of a quantity that could be modelled using the normal distribution.

1b
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3 marks

The graph of the normal distribution has a characteristic bell shape that is symmetrical about the mean, mu.  If X has a normal distribution with a mean,mu , and variance,sigma squared,  then it can be written as X tilde space N left parenthesis mu comma sigma squared right parenthesis.

 For X tilde N left parenthesis mu comma sigma squared right parenthesis, state:

(i)
straight P left parenthesis X space less than space mu right parenthesis

(ii)
straight P left parenthesis X space greater or equal than space mu right parenthesis

(iii)
straight P left parenthesis X space equals space mu right parenthesis
1c
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1 mark

Using your answers to part (b), or otherwise, explain why there is no difference between greater or equal than and greater than or less or equal than and less than when calculating normal probabilities.

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2a
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5 marks

For the standard normal distribution, Z tilde straight N open parentheses 0 comma 1 squared close parentheses, the probability straight P open parentheses Z less or equal than z close parentheses can be written as straight capital phi left parenthesis z right parenthesis

The value of  straight capital phi left parenthesis 0.23 right parenthesis spacewould be found in the tables at the point where the ‘0.2’ row and the ‘3’ column meet.  This value is 0.5910. 

Use the table of z-values in the formula booklet to write down the value of:

(i)
straight capital phi left parenthesis 0.60 right parenthesis
(ii)
straight capital phi left parenthesis 0.63 right parenthesis
(iii)
straight capital phi left parenthesis 1 right parenthesis
(iv)
straight capital phi left parenthesis 1.59 right parenthesis
(v)
straight capital phi left parenthesis 2.99 right parenthesis
2b
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5 marks

To find the value of straight capital phi left parenthesis 0.236 right parenthesis from the tables we would need to find how much to add on to the value of 0.5910 for an added probability of 0.006.  To do this we find the value at the point where the ’0.2’ row and the ‘ADD 6’ column meet.  This gives us a value of 23 representing 0.0023.  So, we must find 0.5910 plus 0.0023 space equals space 0.5933, therefore straight capital phi left parenthesis 0.236 right parenthesis equals 0.5933 to four decimal places. 

Use the table of z-values in the formula booklet to calculate the value of:

(i)
straight capital phi left parenthesis 0.601 right parenthesis
(ii)
straight capital phi left parenthesis 0.635 right parenthesis
(iii)
straight capital phi left parenthesis 1.004 right parenthesis
(iv)
straight capital phi left parenthesis 1.598 right parenthesis
(v)
straight capital phi left parenthesis 2.999 right parenthesis

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3a
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4 marks

For the standard normal distribution, Z space tilde space straight N left parenthesis 0 comma 1 ² right parenthesis ,the probability straight P left parenthesis Z space less than space z right parenthesis spacecan be written as  text Φ end text left parenthesis z right parenthesis space.

To solve space straight capital phi left parenthesis straight z right parenthesis equals 0.6808  we would find the value 0.6808 from within the tables and read off the corresponding z-value.   0.6808 is found at the point where the ‘0.4’ row and the ‘7’ column meet, so straight capital phi left parenthesis 0.47 right parenthesis equals 0.6808

To solve straight capital phi left parenthesis straight z right parenthesis equals 0.878 we can use the tables to find straight capital phi left parenthesis 1.16 right parenthesis equals 0.8770, there is a 10 in the ‘ADD 5’ column so straight capital phi left parenthesis 1.165 right parenthesis equals 0.8780

Use the table of -values in the formula booklet to solve:

(i)
straight capital phi left parenthesis straight z right parenthesis equals 0.5398
(ii)
straight capital phi left parenthesis straight z right parenthesis equals 0.9147
(iii)
straight capital phi left parenthesis straight z right parenthesis equals 0.8642
(iv)
straight capital phi left parenthesis straight z right parenthesis equals 0.9764
3b
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3 marks

If a probability does not appear in the table, then we can take the midpoint of the z-values that are either side of the probability.

(i)

Use the table of z -values to find the largest value of a, to three decimal places, such that space straight capital phi left parenthesis straight a right parenthesis less than 0.7.

(ii)
Use the table of z -values to find the smallest value of b, to three decimal places, such that space straight capital phi left parenthesis straight b right parenthesis greater than 0.7.
(iii)
By finding the midpoint of a and b, find an approximate solution to space straight capital phi left parenthesis straight z right parenthesis equals 0.7 to four decimal places.
3c
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2 marks

The formula booklet also contains a table of critical values of z such that straight P open parentheses Z less or equal than z close parentheses equals p. From this table we can see that if space straight P left parenthesis Z less or equal than z right parenthesis equals 0.9995 spacethen space z equals 3.291 space to three decimal places. 

Use the table of critical values to find the value of z such that:

(i)
straight P left parenthesis Z less or equal than z right parenthesis equals 0.75
(ii)
straight P left parenthesis Z less or equal than z right parenthesis equals 0.999

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4a
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4 marks

For the standard normal distribution, Z space tilde space straight N left parenthesis 0 comma 1 ² right parenthesis, the probability  straight P left parenthesis Z less or equal than space z right parenthesis can be written as straight capital phi left parenthesis straight z right parenthesis. 

To calculate straight P left parenthesis Z greater than z right parenthesis  the following formulae can be used:

         straight P left parenthesis straight Z greater than straight z right parenthesis equals 1 minus straight capital phi left parenthesis straight z right parenthesis space space space space space spaceand   space space space straight P left parenthesis straight Z greater than straight z right parenthesis equals straight capital phi left parenthesis negative z right parenthesis

Use the formulae above to write each of the following in the form space 1 minus straight capital phi left parenthesis straight z right parenthesis space spaceor space straight capital phi left parenthesis negative z right parenthesis space spaceand then use the table of straight z-values to find its value:

(i)
straight P left parenthesis straight Z greater than 0.74 right parenthesis
(ii)
straight P left parenthesis Z greater than negative 0.74 right parenthesis
(iii)
straight P left parenthesis Z greater than 2.142 right parenthesis
(iv)
straight P left parenthesis Z greater than negative 1.516 right parenthesis
4b
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5 marks

Use the formulae above to rewrite each equation in the form space straight capital phi left parenthesis z right parenthesis equals p or straight capital phi left parenthesis negative z right parenthesis equals p and then use the table of straight z-values and the table of critical values to find the value of straight z such that:

(i)
straight P left parenthesis straight Z greater than straight z right parenthesis equals 0.9306
(ii)
straight P left parenthesis straight Z greater than straight z right parenthesis equals 0.0694
(iii)
straight P left parenthesis straight Z greater than straight z right parenthesis equals 0.1292
(iv)
straight P left parenthesis straight Z greater than straight z right parenthesis equals 0.5931

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5a
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3 marks

For the standard normal distribution, Z space tilde space straight N left parenthesis 0 comma 1 ² right parenthesis, the probability straight P left parenthesis straight Z less or equal than space straight z right parenthesis can be written as straight capital phi left parenthesis straight z right parenthesis.

 The table of straight z-values does not contain negative straight z-values. To calculate probabilities involving negative straight z-values the following formula can be used:

               straight capital phi left parenthesis negative straight z right parenthesis equals 1 minus straight capital phi left parenthesis straight z right parenthesis.

Use the table of straight z-values in the formula booklet and the formula above to find the value of:

(i)
straight capital phi left parenthesis negative 0.5 right parenthesis
(ii)
straight capital phi left parenthesis negative 0.579 right parenthesis
(iii)
straight capital phi left parenthesis negative 2.785 right parenthesis

 

5b
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5 marks

Use the formula above to find the value of space straight capital phi left parenthesis straight z right parenthesis space spaceand then use the table of straight z minusvalues and the table of critical values in the formula booklet to find the value of negative straight z and hence solve:

(i)
straight capital phi left parenthesis straight z right parenthesis equals 0.1093
(ii)
straight capital phi left parenthesis straight z right parenthesis equals 0.001
(iii)
straight capital phi left parenthesis straight z right parenthesis equals 0.0549
(iv)
straight capital phi left parenthesis straight z right parenthesis equals 0.0005

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6a
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5 marks

For the standard normal distribution, Z space tilde space straight N left parenthesis 0 comma 1 ² right parenthesis, the probability  straight P left parenthesis Z less or equal than space z right parenthesis can be written as straight capital phi left parenthesis straight z right parenthesis

To calculate the probability between two values the following formula can be used:.

               straight P left parenthesis a less than Z less than b right parenthesis equals straight capital phi left parenthesis b right parenthesis minus straight capital phi left parenthesis a right parenthesis

Use the formula above to rewrite the following in the form space straight capital phi left parenthesis b right parenthesis minus straight capital phi left parenthesis a right parenthesis space spaceand then use the table of straight z-values to calculate the probability and round your answer to three decimal places:

(i)
straight P left parenthesis 0.12 less than Z less than 1.34 right parenthesis
(ii)
straight P left parenthesis 1.037 less than Z less than 2.913 right parenthesis
(iii)
straight P left parenthesis negative 2 less than Z less than 1 right parenthesis
(iv)
straight P left parenthesis negative 1.5 less than Z less than negative 0.5 right parenthesis
6b
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6 marks

Use the formula above to rewrite the following in the form space straight capital phi left parenthesis b right parenthesis minus straight capital phi left parenthesis a right parenthesis equals p spaceand then use the table of straight z-values and the table of critical values to find the value of straight z such that:

(i)
straight P left parenthesis 0 less than Z less than z right parenthesis equals 0.499
(ii)
straight P left parenthesis z less than Z less than 2.093 right parenthesis equals 0.319
(iii)
straight P left parenthesis z less than Z less than 1.3 right parenthesis equals 0.9027
(iv)
straight P left parenthesis negative z less than Z less than z right parenthesis equals 0.6826

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7a
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4 marks

For the standard normal distribution, Z space tilde space straight N left parenthesis 0 comma 1 ² right parenthesis, the probability  straight P left parenthesis Z space less than space z right parenthesis can be written as straight capital phi left parenthesis z right parenthesis

A random variable X space tilde straight N left parenthesis mu comma sigma squared right parenthesis  can be coded to model the standard normal variable Z tilde straight N left parenthesis 0 comma 1 squared right parenthesis comma using the formula:

             space Z equals fraction numerator left parenthesis X minus mu right parenthesis over denominator sigma end fraction

For the random variable X space tilde straight N left parenthesis 21 comma 4 squared right parenthesis,

(i)

write down the values of mu and sigma,

(ii)

calculate the straight z-value that corresponds to space X equals 26

(iii)

write space straight P left parenthesis X greater than 26 right parenthesis in the form straight capital phi left parenthesis straight z right parenthesis,

(iv)
calculate space straight P left parenthesis X greater than 26 right parenthesis.
7b
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6 marks

For the random variable X tilde space straight N left parenthesis 54 comma 5 squared right parenthesis,  write in terms of straight capital phi left parenthesis z right parenthesis and hence find:

(i)
straight P left parenthesis X less or equal than 60 right parenthesis
(ii)
straight P open parentheses X less than 51 close parentheses
(iii)
straight P open parentheses X greater or equal than 58 close parentheses.
7c
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4 marks

For space X space tilde space straight N left parenthesis 187 comma 100 right parenthesis,

(i)
explain why space sigma equals 10,
(ii)

write space straight P left parenthesis 190 less than X less than 203 right parenthesis space spacein the form space straight P left parenthesis a less than Z less than b right parenthesis

(iii)
hence calculate space straight P left parenthesis 190 less than X less than 203 right parenthesis.

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1a
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2 marks

For the random variable, space space X tilde straight N left parenthesis 32 comma 9 right parenthesis,

(i)
Write down the mean and standard deviation.
(ii)
Draw a sketch of the graph and clearly label the mean.  
1b
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2 marks
(i)
With the help of your diagram, explain how you should know, without carrying out any calculations, that  straight P left parenthesis X less or equal than 34 right parenthesis greater than 0.5.
(ii)
With the help of your diagram, explain how you should know, without carrying out any calculations, that straight P left parenthesis X greater or equal than 39 right parenthesis less than 0.5.
1c
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6 marks

Use the formula  Z equals fraction numerator X minus mu over denominator sigma end fraction  to find appropriate -values, to three decimal places, and then use the table of straight z-values to calculate, to three decimal places,

(i)
straight P left parenthesis X less or equal than space 34 right parenthesis
(ii)
straight P left parenthesis X greater or equal than 39 right parenthesis
(iii)
P open parentheses 34 less or equal than X less or equal than 39 close parentheses.

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2a
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4 marks

For the standard normal distribution Z tilde straight N left parenthesis 0 comma 1 squared right parenthesis, use the table of straight z-values to find:

(i)
straight P left parenthesis Z less than 1.5 right parenthesis
(ii)
straight P left parenthesis Z greater than negative 0.8 right parenthesis
(iii)
straight P left parenthesis negative 2.1 less than Z less than negative 0.3 right parenthesis
2b
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3 marks

The random variable X tilde straight N left parenthesis 2 comma 0.1 squared right parenthesis. 

Using the coding relationship between X and Z, find the values of a, b, c and d such that:

(i)
straight P left parenthesis X less than a right parenthesis space equals space straight P left parenthesis Z less than 1.5 right parenthesis
(ii)
straight P left parenthesis X greater than b right parenthesis space equals space straight P left parenthesis Z greater than negative 0.8 right parenthesis
(iii)
straight P left parenthesis c less than X less than d right parenthesis space equals space straight P left parenthesis negative 2.1 less than Z less than negative 0.3 right parenthesis

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3a
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3 marks

The random variable X tilde straight N left parenthesis 50 comma 49 right parenthesis

(i)
Write down the values of mu and sigma .
(ii)
Use the table of critical values to find the value of straight z such that straight P left parenthesis Z less or equal than z right parenthesis space equals space 0.975 space.

(iii)    Use the formula Z equals fraction numerator X minus mu over denominator sigma end fraction to find the value of a such that straight P left parenthesis X less or equal than a right parenthesis space equals space 0.975.

3b
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2 marks
(i)
Find the value of straight z such that straight P left parenthesis Z greater or equal than z right parenthesis space equals space 0.001.

(ii)     Use the formula Z equals fraction numerator X minus mu over denominator sigma end fraction to find the value of b such that space straight P left parenthesis X greater or equal than b right parenthesis equals 0.001.

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4a
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4 marks

For the random variable space X space tilde space straight N left parenthesis mu comma 36 right parenthesis it is known that space straight P left parenthesis X less than 51 right parenthesis equals 0.8665.

(i)
Write down the value of sigma.
(ii)
Calculate the value of straight z such that space straight P left parenthesis Z less than z right parenthesis equals 0.8665.

(iii)     Use the formula Z equals fraction numerator X minus mu over denominator sigma end fraction to show that space 51 minus mu equals 6.66.

(iv)
Hence calculate the value of mu.
4b
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3 marks

For the random variable space Y space tilde space straight N left parenthesis 93 comma sigma squared right parenthesis it is known that space straight P left parenthesis Y greater than 101 right parenthesis equals 0.0735.

(i)
Calculate the value of straight z such that space straight P left parenthesis Z greater than z right parenthesis equals 0.0735.

(ii)     Use the formula Z equals fraction numerator X minus mu over denominator sigma end fraction to show that space 1.45 sigma equals 8.

(iii)
Hence calculate the value of sigma.

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5a
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5 marks

For the random variable X tilde straight N left parenthesis 23 comma 4 squared right parenthesis find the following probabilities:

(i)
straight P left parenthesis X less than 20 right parenthesis
(ii)
straight P left parenthesis X greater or equal than 29 right parenthesis
(iii)
 straight P left parenthesis 20 less or equal than X less than 29 right parenthesis

 

5b
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5 marks

For the random variable Y tilde straight N left parenthesis 100 comma 225 right parenthesis find the following probabilities:

(i)
straight P left parenthesis Y less or equal than 90 right parenthesis
(ii)
straight P left parenthesis Y greater than 140 right parenthesis
(iii)
straight P left parenthesis 85 less or equal than Y less or equal than 115 right parenthesis

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6a
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5 marks

The random variable X tilde straight N left parenthesis 10 comma 9 right parenthesis.

Find the value of a and the value of b, each to 2 decimal places, such that:

(i)
straight P left parenthesis X less than a right parenthesis equals 0.4
(ii)
straight P left parenthesis X greater than b right parenthesis equals 0.25

 

6b
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2 marks

Use a sketch of the distribution of X to explain why straight P left parenthesis a less or equal than X less or equal than b right parenthesis equals 0.35.

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7a
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2 marks

The random variable X tilde N left parenthesis mu comma sigma squared right parenthesis. It is known that straight P left parenthesis X greater than 36.88 right parenthesis equals 0.025 and straight P left parenthesis X less than 27.16 right parenthesis equals 0.1

Find the values of a and b for which space straight P left parenthesis Z greater than a right parenthesis equals 0.025 and straight P left parenthesis Z less than b right parenthesis equals 0.1 comma space spacewhere Z is the standard normal variable.  Give your answers correct to 3 decimal places.

7b
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2 marks

Use your answers to part (a), along with the relationship between Z and X, to show that the following simultaneous equations must be true:

                  mu plus 1.96 sigma equals 36.88

                  mu minus 1.2816 sigma equals 27.16

7c
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2 marks

By solving the simultaneous equations in (b), determine the values of mu and sigma. Give your answers correct to 2 decimal places.

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1a
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5 marks

The test scores, X, of a group of RAF recruits in an aptitude test are modelled as a normal distribution with X tilde straight N left parenthesis 210 comma 27.8 squared right parenthesis.

(i)
Find the values of a and b such that  straight P left parenthesis X less than a right parenthesis equals 0.25 and straight P left parenthesis X greater than b right parenthesis equals 0.25.
(ii)
Hence find the interquartile range of the scores.
1b
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3 marks

Those who score in the top 30% on the test move on to the next stage of training.

One of the recruits, Amelia, achieves a score of 231. Determine whether Amelia will move on to the next stage of training.

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2a
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8 marks

The random variable  X tilde straight N left parenthesis 13 comma 15 right parenthesis.

Find the value of a, to 3 significant figures, such that:

(i)
straight P left parenthesis X greater than a right parenthesis equals 0.2
(ii)
straight P left parenthesis a less or equal than X less or equal than 14 right parenthesis equals 0.5
(iii)
straight P left parenthesis 13 minus a less or equal than X less or equal than 13 plus a right parenthesis equals 0.9372
2b
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2 marks

Explain why there are no values of a such that space straight P left parenthesis 14 less or equal than X less or equal than a right parenthesis equals 0.5.

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3a
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3 marks

The standard normal distribution is Z space tilde straight N left parenthesis 0 comma 1 squared right parenthesis.

(i)
Find the value of straight z for which straight P left parenthesis Z less than z right parenthesis equals 0.9.
(ii)
Use your answer to part (a)(i) along with the properties of the normal distribution to work out the values of  a and b for which straight P left parenthesis Z greater than a right parenthesis equals 0.1 and straight P left parenthesis Z greater than b right parenthesis equals 0.9.
3b
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2 marks

The weights, W k g comma of coconuts grown on the Coconutty As They Come coconut plantation are modelled as a normal distribution with mean 1.25 kg and standard deviation 0.38 kg.  The plantation only considers coconuts to be exportable if their weights are greater than 10% but less than 90%. 

Use your answer to part (a)(ii) to find the range of possible weights, to the nearest 0.01 k g, for an exportable coconut.

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4a
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4 marks

The random variable X tilde straight N left parenthesis mu comma sigma squared right parenthesis. It is known that  straight P left parenthesis X greater than 34.451 right parenthesis equals 0.001 and straight P left parenthesis X less than 14.792 right parenthesis equals 0.1966.

(i)
Use the relationship between X and the standard normal variable Z to show that the following equation must be true:

            mu plus 3.090 sigma equals 34.451 space
(ii)
Write down a second equation in terms of mu and  sigma.
4b
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2 marks

By solving the simultaneous equations in (a), determine the values of mu and sigma

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5a
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3 marks

A machine is used to fill bottles of a particular brand of soft drink.  The volume, V ml, of soft drink in the bottles is normally distributed with mean 450 ml and standard deviation sigma ml.  Given that 5% of the cans contain less than 429 ml of soft drink, find: 

The value of sigma,

5b
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3 marks

 straight P left parenthesis V greater or equal than 485 right parenthesis correct to 3 decimal places,

5c
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2 marks

The probability that the volume of a randomly selected bottle is no more than one standard deviation away from the mean.

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6a
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3 marks

Paul enjoys solving sudoku puzzles.  The lengths of time he spends on sudokus in a week are normally distributed with a mean of 2048 minutes and a standard deviation of 64 minutes. 

Find the probability that in a given week Paul spends less than 1945 minutes solving sudoku puzzles.

6b
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4 marks

Estimate the number of weeks in a year that Paul spends between 2019 and 2091 minutes solving sudoku puzzles.

6c
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4 marks

Assuming it takes Paul exactly 10 minutes to solve any sudoku puzzle, find the greatest number, n, of sudoku puzzle such that the probability of Paul solving less than n puzzles in a week is less than 0.01.

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7a
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5 marks

A machine is used to fill bags of potatoes for a supermarket chain.  The weight, W kg, of potatoes in the bags is normally distributed with mean 3 k g and standard deviation sigma k g

Given that 7% of the bags contain a weight of potatoes that is at least 50 g more than  the mean, find:

         straight P left parenthesis 2.9 less or equal than W less or equal than 3.1 right parenthesis.

7b
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4 marks

Twelve of the bags of potatoes are chosen at random.

Find the probability that not more than one of the bags will contain less than 2.96 kg of potatoes.

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1a
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4 marks

The test scores, X, of a group of Royal Navy recruits in an aptitude test are modelled as a normal distribution with X tilde straight N left parenthesis 520 comma 89.9 squared right parenthesis.

Find the interquartile range of the scores.

1b
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5 marks

Those who get a score over 670 are offered a role within the submarine service.  Only those who get a score over 750 are offered a leadership role within the submarine service. 

Given that Mervyn, one of the recruits, has been offered a role within the submarine service, find the probability that he is offered a leadership role.

 

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2a
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4 marks

The distribution of the test scores, X, of a group of British Army officer cadets on an aviation aptitude test is modelled as a normal distribution with  X tilde N left parenthesis 120 comma 26.5 squared right parenthesis

Only cadets who score in the top 10% on the test are eligible to proceed directly to helicopter pilot training.  Cadets whose scores are between the 40th and 90th percentiles, however, are eligible to resit the test in an attempt to improve their scores. 

Given that it is only possible to receive an integer number of marks as a score on the test, determine the range of test scores for which cadets would be eligible to resit the test.

2b
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4 marks
(i)
Find straight P left parenthesis X greater than 180 right parenthesis.
(ii)

The maximum score it is possible to receive on the test is 180.  Use this fact, and your answer to part (b)(i), to criticise the model being used for the score distribution.

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3
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6 marks

The weight, W kg, of the feed in a sack of pheasant feed produced by a certain manufacturer is modelled as  W tilde N open parentheses table row cell 20 comma end cell cell 1 over 3600 end cell end table close parentheses .

Roger buys twelve sacks of the manufacturer’s pheasant feed to feed to the pheasants who have begun showing up at his backyard bird feeding station.

Find the probability that all twelve sacks contain feed with a weight that is within 35 straight g of 20 kg.

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4a
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6 marks

The masses of turkeys can be modelled using a normal distribution with a mean of 4.7 space kg and a variance of 1.9 space kg ². Nicholas, a farmer, classes a turkey as ‘holiday ready’ if it weighs more than 5.5 space kg

A turkey is selected at random. Given that it is ‘holiday ready’, find the probability that it weighs less than 6 space kg.

4b
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5 marks

Nicholas takes a large sample of ‘holiday ready’ turkeys, estimate the median mass of the turkeys in his sample.

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5a
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6 marks

An archaeologist has devoted his life to studying ancient Greek vases produced by a particular Boeotian pottery workshop.  The vases were made to a standard pattern, and after measuring a very large number of them the archaeologist has found that 5% of the vases have a mass greater than 2.237 space kg, while only 1% of them have a mass less than 1.906 space kg

Given that the masses of the vases may be assumed to be distributed normally, find the mean and standard deviation of the distribution.

5b
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4 marks

The archaeologist has found that vases made by the workshop with a mass less than m space kg are particularly fragile and require special care. A museum has just purchased a collection of  vases produced by the workshop. The probability of all 10 vases requiring special care is 9.766 cross times 10 to the power of negative 24 end exponent. 

Find the value of m.

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6a
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5 marks

The Monkey Puzzle Tree Marketing Board has proposed a new scheme that will measure the puzzlingness of monkey puzzle trees in terms of a unit called the ‘fuddle’.  It is found that the puzzlingness measurements, X, of monkey puzzle trees grown on the We Puzzling Monkeys monkey puzzle tree plantation can be modelled as a normal distribution with mean mu fuddles and standard deviation sigma fuddles.  Because the owners of the plantation are committed to making sure that the monkey puzzle trees they sell to gardeners are neither too puzzling nor not puzzling enough, the plantation only considers monkey puzzle trees to be saleable if their puzzlingness is between the space 10 percent sign space and space 97.5 percent sign spacepercentiles of the plantation’s monkey puzzle tree puzzlingness measurements. 

Find the range of possible puzzlingness measurements for a saleable monkey puzzle tree. Your answer should be given in terms of mu and sigma.

6b
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3 marks

Given that space straight P left parenthesis X less than mu minus 8 right parenthesis equals 0.2 comma find space straight P left parenthesis X greater than mu plus 8 │ X greater than mu minus 8 right parenthesis.

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