Normal Distribution (Cambridge (CIE) A Level Maths: Probability & Statistics 1): Exam Questions

Exam code: 9709

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, μ.  If X has a normal distribution with a mean,μ , and variance,σ2,  then it can be written as X~ N(μ,σ2).

 For X~N(μ,σ2), state:

(i) P(X < μ)

(ii) P(X  μ)

(iii) P(X = μ)

1c
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1 mark

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

2a
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5 marks

For the standard normal distribution, Z~N(0,12), the probability P(Zz) can be written as Φ(z)

The value of  Φ(0.23) would 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) Φ(0.60)

(ii) Φ(0.63)

(iii) Φ(1)

(iv) Φ(1.59)

(v) Φ(2.99)

2b
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5 marks

To find the value of Φ(0.236) 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+0.0023 = 0.5933, therefore Φ(0.236)=0.5933 to four decimal places. 

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

(i) Φ(0.601)

(ii) Φ(0.635)

(iii) Φ(1.004)

(iv) Φ(1.598)

(v) Φ(2.999)

3a
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4 marks

For the standard normal distribution, Z ~ N(0,1²) ,the probability P(Z < z) can be written as  Φ(z) .

To solve  Φ(z)=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 Φ(0.47)=0.6808

To solve Φ(z)=0.878 we can use the tables to find Φ(1.16)=0.8770, there is a 10 in the ‘ADD 5’ column so Φ(1.165)=0.8780

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

(i) Φ(z)=0.5398

(ii) Φ(z)=0.9147

(iii) Φ(z)=0.8642

(iv) Φ(z)=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  Φ(a)<0.7.

(ii) Use the table of z -values to find the smallest value of b, to three decimal places, such that  Φ(b)>0.7.

(iii) By finding the midpoint of a and b, find an approximate solution to  Φ(z)=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 P(Zz)=p. From this table we can see that if  P(Zz)=0.9995 then  z=3.291  to three decimal places. 

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

(i) P(Zz)=0.75

(ii) P(Zz)=0.999

4a
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4 marks

For the standard normal distribution, Z ~ N(0,1²), the probability  P(Z z) can be written as Φ(z)

To calculate P(Z>z)  the following formulae can be used:

         P(Z>z)=1Φ(z)      and      P(Z>z)=Φ(z)

Use the formulae above to write each of the following in the form  1Φ(z)  or  Φ(z)  and then use the table of z-values to find its value:

(i) P(Z>0.74)

(ii) P(Z>0.74)

(iii) P(Z>2.142)

(iv) P(Z>1.516)

4b
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5 marks

Use the formulae above to rewrite each equation in the form  Φ(z)=p or Φ(z)=p and then use the table of z-values and the table of critical values to find the value of z such that:

(i) P(Z>z)=0.9306

(ii) P(Z>z)=0.0694

(iii) P(Z>z)=0.1292

(iv) P(Z>z)=0.5931

5a
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3 marks

For the standard normal distribution, Z ~ N(0,1²), the probability P(Z z) can be written as Φ(z).

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

               Φ(z)=1Φ(z).

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

(i) Φ(0.5)

(ii) Φ(0.579)

(iii) Φ(2.785)

 

5b
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5 marks

Use the formula above to find the value of  Φ(z)  and then use the table of zvalues and the table of critical values in the formula booklet to find the value of z and hence solve:

(i) Φ(z)=0.1093

(ii) Φ(z)=0.001

(iii) Φ(z)=0.0549

(iv) Φ(z)=0.0005

6a
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5 marks

For the standard normal distribution, Z ~ N(0,1²), the probability  P(Z z) can be written as Φ(z)

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

               P(a<Z<b)=Φ(b)Φ(a)

Use the formula above to rewrite the following in the form  Φ(b)Φ(a)  and then use the table of z-values to calculate the probability and round your answer to three decimal places:

(i) P(0.12<Z<1.34)

(ii) P(1.037<Z<2.913)

(iii) P(2<Z<1)

(iv) P(1.5<Z<0.5)

6b
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6 marks

Use the formula above to rewrite the following in the form  Φ(b)Φ(a)=p and then use the table of z-values and the table of critical values to find the value of z such that:

(i) P(0<Z<z)=0.499

(ii) P(z<Z<2.093)=0.319

(iii) P(z<Z<1.3)=0.9027

(iv) P(z<Z<z)=0.6826

7a
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4 marks

For the standard normal distribution, Z ~ N(0,1²), the probability  P(Z < z) can be written as Φ(z)

A random variable X ~N(μ,σ2)  can be coded to model the standard normal variable Z~N(0,12), using the formula:

              Z=(Xμ)σ

For the random variable X ~N(21,42),

(i) write down the values of μ and σ,

(ii) calculate the z-value that corresponds to  X=26

(iii) write  P(X>26) in the form Φ(z),

(iv) calculate  P(X>26).

7b
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6 marks

For the random variable X~ N(54,52),  write in terms of Φ(z) and hence find:

(i) P(X60)

(ii) P(X<51)

(iii) P(X58).

7c
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4 marks

For  X ~ N(187,100),

(i) explain why  σ=10,

(ii) write  P(190<X<203)  in the form  P(a<Z<b)

(iii) hence calculate  P(190<X<203).

1a
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2 marks

For the random variable,   X~N(32,9),

(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  P(X34)>0.5.

(ii) With the help of your diagram, explain how you should know, without carrying out any calculations, that P(X39)<0.5.

1c
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6 marks

Use the formula  Z=Xμσ  to find appropriate -values, to three decimal places, and then use the table of z-values to calculate, to three decimal places,

(i) P(X 34)

(ii) P(X39)

(iii) P(34X39).

2a
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4 marks

For the standard normal distribution Z~N(0,12), use the table of z-values to find:

(i) P(Z<1.5)

(ii) P(Z>0.8)

(iii) P(2.1<Z<0.3)

2b
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3 marks

The random variable X~N(2,0.12). 

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

(i) P(X<a) = P(Z<1.5)

(ii) P(X>b) = P(Z>0.8)

(iii) P(c<X<d) = P(2.1<Z<0.3)

3a
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3 marks

The random variable X~N(50,49)

(i) Write down the values of μ and σ .

(ii)Use the table of critical values to find the value of z such that P(Zz) = 0.975 .

(iii)Use the formula Z=Xμσ to find the value of a such that P(Xa) = 0.975.

3b
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2 marks

(i) Find the value of z such that P(Zz) = 0.001.

(ii) Use the formula Z=Xμσ to find the value of b such that  P(Xb)=0.001.

4a
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4 marks

For the random variable  X ~ N(μ,36) it is known that  P(X<51)=0.8665.

(i) Write down the value of σ.

(ii)Calculate the value of z such that  P(Z<z)=0.8665.

(iii)Use the formula Z=Xμσ to show that  51μ=6.66.

(iv) Hence calculate the value of μ.

4b
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3 marks

For the random variable  Y ~ N(93,σ2) it is known that  P(Y>101)=0.0735.

(i) Calculate the value of z such that  P(Z>z)=0.0735.

(ii) Use the formula Z=Xμσ to show that  1.45σ=8.

(iii) Hence calculate the value of σ.

5a
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5 marks

For the random variable X~N(23,42) find the following probabilities:

(i) P(X<20)

(ii) P(X29)

(iii)  P(20X<29)

 

5b
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5 marks

For the random variable Y~N(100,225) find the following probabilities:

(i) P(Y90)

(ii) P(Y>140)

(iii) P(85Y115)

6a
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5 marks

The random variable X~N(10,9).

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

(i) P(X<a)=0.4

(ii) P(X>b)=0.25

 

6b
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2 marks

Use a sketch of the distribution of X to explain why P(aXb)=0.35.

7a
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2 marks

The random variable X~N(μ,σ2). It is known that P(X>36.88)=0.025 and P(X<27.16)=0.1

Find the values of a and b for which  P(Z>a)=0.025 and P(Z<b)=0.1,  where 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:

                  μ+1.96σ=36.88

                  μ1.2816σ=27.16

7c
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2 marks

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

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~N(210,27.82).

(i) Find the values of a and b such that  P(X<a)=0.25 and P(X>b)=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.

2a
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8 marks

The random variable  X~N(13,15).

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

(i) P(X>a)=0.2

(ii) P(aX14)=0.5

(iii) P(13aX13+a)=0.9372

2b
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2 marks

Explain why there are no values of a such that  P(14Xa)=0.5.

3a
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3 marks

The standard normal distribution is Z ~N(0,12).

(i) Find the value of z for which P(Z<z)=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 P(Z>a)=0.1 and P(Z>b)=0.9.

3b
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2 marks

The weights, W kg, 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 kg, for an exportable coconut.

4a
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4 marks

The random variable X~N(μ,σ2). It is known that  P(X>34.451)=0.001 and P(X<14.792)=0.1966.

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

            μ+3.090σ=34.451 

(ii) Write down a second equation in terms of μ and  σ.

4b
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2 marks

By solving the simultaneous equations in (a), determine the values of μ and σ

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 σ ml.  Given that 5% of the cans contain less than 429 ml of soft drink, find: 

The value of σ,

5b
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3 marks

 P(V485) 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.

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.

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 kg and standard deviation σ kg

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

          P(2.9W3.1).

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.

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~N(520,89.92).

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.

 

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~N(120,26.52)

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 P(X>180).

(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.

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~N(20,13600) .

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 g of 20 kg.

4a
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6 marks

The masses of turkeys can be modelled using a normal distribution with a mean of 4.7 kg and a variance of 1.9 kg². Nicholas, a farmer, classes a turkey as ‘holiday ready’ if it weighs more than 5.5 kg

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

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 kg, while only 1% of them have a mass less than 1.906 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 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 9.766×1024. 

Find the value of m.

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 μ fuddles and standard deviation σ 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  10% and 97.5% percentiles 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 μ and σ.

6b
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3 marks

Given that  P(X<μ8)=0.2, find  P(X>μ+8X>μ8).