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ZIMSEC A Level · 9164/4 · J2014

Statistics Paper 4 June 2014

Questions
43
Total marks
96
Syllabus code
9164/4

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Questions
43
Pass mark
26
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Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Section a

Section a, Question 2

[2 marks]Probability without replacement

A bag contains 24 counters of which 6 are red, 8 are green and 10 are yellow. Three counters are taken from the bag at random without replacement.

Which calculation gives the probability that exactly 2 of the three counters taken are green?

  1. A(82)(161)(242)\dfrac{\binom{8}{2}\binom{16}{1}}{\binom{24}{2}}, choosing 2 from the 8 green ones and 1 from the 16 that are not green, over pairs drawn from 24
  2. B(82)(243)\dfrac{\binom{8}{2}}{\binom{24}{3}}, choosing 2 counters from the 8 green ones and leaving the third counter out of the count altogether
  3. C(82)(221)(243)\dfrac{\binom{8}{2}\binom{22}{1}}{\binom{24}{3}}, choosing 2 from the 8 green ones and 1 from the 22 counters that remain once two have gone
  4. D(82)(161)(243)\dfrac{\binom{8}{2}\binom{16}{1}}{\binom{24}{3}}, choosing 2 from the 8 green ones and 1 from the 16 counters that are not green
[3 marks]Probability without replacement

A bag contains 24 counters of which 6 are red, 8 are green and 10 are yellow. Three counters are taken from the bag at random without replacement. The probability that exactly 2 of the counters taken are green is 56253\dfrac{56}{253}.

Given that 2 of the counters are green, find the probability that the first counter taken is red.

Answer this when you sit the paper.

[1 marks]Geometric distribution

The probability that a boy hits a target is 0.8. Shots are independent of each other, and during each practice period the boy fires shots until he hits the target.

Find the mean number of shots fired per practice period.

Answer this when you sit the paper.

[2 marks]Geometric distribution

The probability that a boy hits a target is 0.8. Shots are independent of each other, and during each practice period the boy fires shots until he hits the target.

Find the standard deviation of the number of shots fired per practice period.

Answer this when you sit the paper.

[2 marks]Geometric distribution

The probability that a boy hits a target is 0.8. Shots are independent of each other, and during each practice period the boy fires shots until he hits the target.

Find the probability that the boy will need to take at least five shots to hit the target.

Answer this when you sit the paper.

[3 marks]Binomial hypothesis test

A coin is tossed 5 times and at least 4 heads are obtained. A binomial test is used at the 5% level of significance to test whether the coin is biased towards heads, with H0:p=0.5H_{0}: p = 0.5 against H1:p>0.5H_{1}: p > 0.5, so that under H0H_{0} the number of heads X has the distribution B(5, 0.5)B(5,\ 0.5).

Find P(X≥4)P(X \ge 4), the probability of the observed result or a more extreme one under H0H_{0}.

Answer this when you sit the paper.

[2 marks]Binomial hypothesis test

A coin is tossed 5 times and at least 4 heads are obtained. A binomial test is used at the 5% level of significance to test whether the coin is biased towards heads, with H0:p=0.5H_{0}: p = 0.5 against H1:p>0.5H_{1}: p > 0.5, so that under H0H_{0} the number of heads X has the distribution B(5, 0.5)B(5,\ 0.5). The tail probability works out as P(X≥4)=0.1875P(X \ge 4) = 0.1875.

What is the conclusion of the test?

  1. A0.18750.1875 is greater than 0.050.05, so the result is not in the critical region, H0H_{0} is not rejected and there is no evidence that the coin is biased towards heads.
  2. B0.18750.1875 is greater than 0.050.05, so the result is in the critical region, H0H_{0} is rejected and there is evidence that the coin is biased towards heads.
  3. C0.18750.1875 is less than 0.950.95, so the result is in the critical region, H0H_{0} is rejected and there is evidence that the coin is biased towards heads.
  4. DP(X=4)=0.15625P(X = 4) = 0.15625 is the figure to compare, and being above 0.050.05 it puts the result in the critical region and rejects H0H_{0}.
[2 marks]Stem and leaf diagrams, quartiles and box plots

The following are television prices in dollars taken in 40 different shops.

40130170240360520170130
240360520120220170330480
160290200120480160210330
7014018026037090150200
28045080420190140270120

Find the median price, in dollars.

Answer this when you sit the paper.

[2 marks]Stem and leaf diagrams, quartiles and box plots

The following are television prices in dollars taken in 40 different shops.

40130170240360520170130
240360520120220170330480
160290200120480160210330
7014018026037090150200
28045080420190140270120

Find the lower quartile of the prices, in dollars.

Answer this when you sit the paper.

[2 marks]Stem and leaf diagrams, quartiles and box plots

The following are television prices in dollars taken in 40 different shops.

40130170240360520170130
240360520120220170330480
160290200120480160210330
7014018026037090150200
28045080420190140270120

Find the upper quartile of the prices, in dollars.

Answer this when you sit the paper.

[1 marks]Stem and leaf diagrams, quartiles and box plots

The following are television prices in dollars taken in 40 different shops.

40130170240360520170130
240360520120220170330480
160290200120480160210330
7014018026037090150200
28045080420190140270120

The lower quartile of these prices is 140 dollars and the upper quartile is 330 dollars. Find the interquartile range, in dollars.

Answer this when you sit the paper.

Section a, Question 3

[3 marks]Normal approximation to the binomial

The probability that a seed grown under specified conditions will germinate and produce a plant is 0.8. A number n of seeds are planted under these conditions and X is the number of them that germinate, so that X∼B(n, 0.8)X \sim B(n,\ 0.8).

Which normal distribution is the suitable approximation to X for large n?

  1. AN(0.8n, 0.16n)N(0.8n,\ 0.16n), using npnp for the mean and npqnpq for the variance of the count
  2. BN(0.8n, 0.8n)N(0.8n,\ 0.8n), setting the variance equal to the mean in the way a Poisson approximation would
  3. CN(0.8n, 0.16n)N(0.8n,\ \sqrt{0.16n}), quoting the standard deviation in the place where the variance belongs
  4. DN(0.2n, 0.16n)N(0.2n,\ 0.16n), using the probability that a seed fails as the mean number that succeed
[2 marks]Normal approximation to the binomial

The probability that a seed grown under specified conditions will germinate and produce a plant is 0.8. A number n of seeds are planted under these conditions and X is the number of them that germinate, so that X∼B(n, 0.8)X \sim B(n,\ 0.8). Requiring a probability of at least 0.9 that 60 or more seeds germinate leads to the inequality n2−149.2n+5531.6≥0n^{2} - 149.2n + 5531.6 \ge 0.

Solve n2−149.2n+5531.6=0n^{2} - 149.2n + 5531.6 = 0 and give the larger of the two roots, correct to 1 decimal place.

Answer this when you sit the paper.

[2 marks]Normal approximation to the binomial

The probability that a seed grown under specified conditions will germinate and produce a plant is 0.8. A number n of seeds are planted under these conditions and X is the number of them that germinate, so that X∼B(n, 0.8)X \sim B(n,\ 0.8). Requiring a probability of at least 0.9 that 60 or more seeds germinate leads to n2−149.2n+5531.6≥0n^{2} - 149.2n + 5531.6 \ge 0, whose roots are n=68.8n = 68.8 and n=80.4n = 80.4.

Find the minimum value of n, the number of seeds to be planted.

Answer this when you sit the paper.

[2 marks]t-test for a population mean

A random sample of eight observations of a normal variable gave xˉ=4.65\bar{x} = 4.65 and ∑(x−xˉ)2=0.74\sum (x - \bar{x})^{2} = 0.74.

Calculate the unbiased estimate of the population variance.

Answer this when you sit the paper.

[3 marks]t-test for a population mean

A random sample of eight observations of a normal variable gave xˉ=4.65\bar{x} = 4.65 and ∑(x−xˉ)2=0.74\sum (x - \bar{x})^{2} = 0.74. The unbiased estimate of the population variance is s2=0.747=0.10571s^{2} = \dfrac{0.74}{7} = 0.10571.

Testing H0:μ=4.32H_{0}: \mu = 4.32 against H1:μ≠4.32H_{1}: \mu \ne 4.32, calculate the value of the test statistic t.

Answer this when you sit the paper.

[2 marks]t-test for a population mean

A random sample of eight observations of a normal variable gave xˉ=4.65\bar{x} = 4.65 and ∑(x−xˉ)2=0.74\sum (x - \bar{x})^{2} = 0.74. Testing H0:μ=4.32H_{0}: \mu = 4.32 against H1:μ≠4.32H_{1}: \mu \ne 4.32 at the 5% level of significance gives a test statistic of t=2.87t = 2.87.

What is the conclusion of the test?

  1. At=2.87t = 2.87 exceeds the 5% critical value 2.365 on 7 degrees of freedom, so H0H_{0} is rejected and the mean is not 4.32.
  2. Bt=2.87t = 2.87 exceeds the 5% critical value 2.306 on 8 degrees of freedom, so H0H_{0} is rejected and the mean is not 4.32.
  3. Ct=2.87t = 2.87 exceeds the 5% critical value 1.960 taken from the normal table, so H0H_{0} is rejected and the mean is not 4.32.
  4. Dt=2.87t = 2.87 falls short of the 5% critical value 2.365 on 7 degrees of freedom, so H0H_{0} is not rejected and the mean may be taken as 4.32.
[2 marks]Poisson distribution

The switchboard of a small company handles both incoming and outgoing calls. During lunch hour on any day, the numbers of incoming and outgoing calls are independent and have Poisson distributions with parameters 5 and 3 respectively.

Find the probability that during lunch hour of a randomly chosen day there will be exactly 3 outgoing calls.

Answer this when you sit the paper.

[2 marks]Poisson distribution

The switchboard of a small company handles both incoming and outgoing calls. During lunch hour on any day, the numbers of incoming and outgoing calls are independent and have Poisson distributions with parameters 5 and 3 respectively.

Find the probability that during lunch hour of a randomly chosen day there will be at least 6 incoming calls.

Answer this when you sit the paper.

[3 marks]Poisson distribution

The switchboard of a small company handles both incoming and outgoing calls. During lunch hour on any day, the numbers of incoming and outgoing calls are independent and have Poisson distributions with parameters 5 and 3 respectively.

Find the probability that during lunch hour of a randomly chosen day there will be a combined number of 3 calls through the switchboard.

Answer this when you sit the paper.

[2 marks]Distribution of the sample mean

A random variable X is normally distributed with mean 15 and standard deviation 6. A random sample of 40 is chosen and found to have mean Xˉ\bar{X}.

What is the distribution of Xˉ\bar{X}?

  1. AXˉ∼N(15, 36×40)\bar{X} \sim N(15,\ 36 \times 40), the population variance multiplied by the size of the sample
  2. BXˉ∼N(15, 0.9)\bar{X} \sim N(15,\ 0.9), the population variance 36 divided by the sample size 40
  3. CXˉ∼N(0.375, 0.9)\bar{X} \sim N(0.375,\ 0.9), the population mean divided by the sample size as well as the variance
  4. DXˉ∼N(15, 36)\bar{X} \sim N(15,\ 36), the population variance carried across to the sample mean unchanged
[3 marks]Distribution of the sample mean

A random variable X is normally distributed with mean 15 and standard deviation 6. A random sample of 40 is chosen and found to have mean Xˉ\bar{X}.

Find P(Xˉ>16)P(\bar{X} > 16).

Answer this when you sit the paper.

[1 marks]Distribution of the sample mean

In a one-tailed calculation the value z is needed for which the area to the right of z under the standard normal curve is 0.05, that is P(Z>z)=0.05P(Z > z) = 0.05.

Write down the value of z, correct to 3 decimal places.

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[3 marks]Distribution of the sample mean

A random variable X is normally distributed with mean 15 and standard deviation 6. A random sample of size n is chosen and has mean Xˉ\bar{X}.

Find the sample size n such that P(Xˉ>15.5)=0.05P(\bar{X} > 15.5) = 0.05.

Answer this when you sit the paper.

Section a, Question 4

[3 marks]Correlation and regression

The data below summarises the altitude x (in metres) above sea level and the mean air temperature y (in ∘^{\circ}C) for 10 weather stations.

∑x=1076∑x2=211642\sum x = 1076 \qquad \sum x^{2} = 211642

∑y=94.2∑xy=9383.1\sum y = 94.2 \qquad \sum xy = 9383.1

∑y2=894.58\sum y^{2} = 894.58

Calculate the product moment correlation coefficient between x and y.

Answer this when you sit the paper.

[2 marks]Correlation and regression

For 10 weather stations, x is the altitude in metres above sea level and y is the mean air temperature in ∘^{\circ}C. The product moment correlation coefficient between x and y works out as r=−0.91r = -0.91.

Which comment on the relationship between x and y is correct?

  1. AThere is almost no linear relationship: knowing the altitude of a station tells you very little about its mean air temperature.
  2. BThere is a weak negative linear relationship: altitude and mean air temperature pull apart from one another only slightly.
  3. CThere is a strong negative linear relationship: the higher the weather station, the lower its mean air temperature.
  4. DThere is a strong positive linear relationship: the higher the weather station, the higher its mean air temperature.
[2 marks]Correlation and regression

The data below summarises the altitude x (in metres) above sea level and the mean air temperature y (in ∘^{\circ}C) for 10 weather stations.

∑x=1076∑x2=211642\sum x = 1076 \qquad \sum x^{2} = 211642

∑y=94.2∑xy=9383.1\sum y = 94.2 \qquad \sum xy = 9383.1

∑y2=894.58\sum y^{2} = 894.58

Calculate the gradient of the regression line of y on x.

Answer this when you sit the paper.

[2 marks]Correlation and regression

For 10 weather stations, x is the altitude in metres above sea level and y is the mean air temperature in ∘^{\circ}C, and the regression line of y on x is

y=10.265−0.00785x.y = 10.265 - 0.00785x.

Use this line to estimate the mean air temperature, in ∘^{\circ}C, at a place 250 m above sea level.

Answer this when you sit the paper.

[2 marks]Normal distribution, sums and differences

A manufacturer sells heavy and light vehicles. The cost of a light vehicle, in thousands of dollars, is normally distributed with mean 252 and standard deviation 2. The cost of a heavy vehicle, in thousands of dollars, is normally distributed with mean 1 012 and standard deviation 5. Costs are independent of one another.

One vehicle of each type is selected at random. Write X for the cost of the light vehicle and Y for the cost of the heavy vehicle. Find the mean of Y−4XY - 4X, in thousands of dollars.

Answer this when you sit the paper.

[2 marks]Normal distribution, sums and differences

A manufacturer sells heavy and light vehicles. The cost of a light vehicle, in thousands of dollars, is normally distributed with mean 252 and standard deviation 2. The cost of a heavy vehicle, in thousands of dollars, is normally distributed with mean 1 012 and standard deviation 5. Costs are independent of one another.

One vehicle of each type is selected at random. Write X for the cost of the light vehicle and Y for the cost of the heavy vehicle. Find the variance of Y−4XY - 4X.

Answer this when you sit the paper.

[3 marks]Normal distribution, sums and differences

A manufacturer sells heavy and light vehicles. The cost of a light vehicle, in thousands of dollars, is normally distributed with mean 252 and standard deviation 2. The cost of a heavy vehicle, in thousands of dollars, is normally distributed with mean 1 012 and standard deviation 5. Costs are independent of one another.

A vehicle of each type is selected at random. Find the probability that the heavy vehicle costs less than 4 times the light vehicle.

Answer this when you sit the paper.

[3 marks]Normal distribution, sums and differences

A manufacturer sells heavy and light vehicles. The cost of a light vehicle, in thousands of dollars, is normally distributed with mean 252 and standard deviation 2. The cost of a heavy vehicle, in thousands of dollars, is normally distributed with mean 1 012 and standard deviation 5. Costs are independent of one another.

One heavy and four light vehicles are selected at random. Find the probability that the cost of the heavy vehicle is less than the total cost of the four light vehicles.

Answer this when you sit the paper.

[2 marks]Chi-squared test of independence

A policeman claims that the type of accident he attends depends on the colour of the car involved. The table shows the results of 200 accidents he attended.

minorseriousfataltotal
black15232260
white35241170
red20232770
Total707060200

Assuming that the type of accident and the colour of the car are independent, calculate the expected frequency of fatal accidents involving black cars.

Answer this when you sit the paper.

[2 marks]Chi-squared test of independence

A policeman claims that the type of accident he attends depends on the colour of the car involved. The table shows the results of 200 accidents he attended.

minorseriousfataltotal
black15232260
white35241170
red20232770
Total707060200

Assuming that the type of accident and the colour of the car are independent, calculate the expected frequency of minor accidents involving white cars.

Answer this when you sit the paper.

[2 marks]Chi-squared test of independence

A policeman claims that the type of accident he attends depends on the colour of the car involved. The table shows the results of 200 accidents he attended.

minorseriousfataltotal
black15232260
white35241170
red20232770
Total707060200

State the number of degrees of freedom for a chi-squared test of independence on this table.

Answer this when you sit the paper.

[3 marks]Chi-squared test of independence

A policeman claims that the type of accident he attends depends on the colour of the car involved. The table shows the results of 200 accidents he attended.

minorseriousfataltotal
black15232260
white35241170
red20232770
Total707060200

Under the hypothesis that the type of accident and the colour of the car are independent, the expected frequencies are 21, 21 and 18 for black cars, 24.5, 24.5 and 21 for white cars, and 24.5, 24.5 and 21 for red cars, taking minor, serious and fatal in that order.

Calculate the value of the chi-squared test statistic.

Answer this when you sit the paper.

[2 marks]Chi-squared test of independence

A policeman claims that the type of accident he attends depends on the colour of the car involved. The table shows the results of 200 accidents he attended.

minorseriousfataltotal
black15232260
white35241170
red20232770
Total707060200

A chi-squared test of independence on this table gives χ2=14.7\chi^{2} = 14.7 on 4 degrees of freedom. The test is carried out at the 10% level of significance.

What is the conclusion?

  1. A14.714.7 exceeds the 10% critical value 7.779 on 4 degrees of freedom, so the colour of a car has been shown to cause the type of accident.
  2. B14.714.7 exceeds the 10% critical value 7.779 on 4 degrees of freedom, so H0H_{0} is rejected and the data supports the policeman's claim.
  3. C14.714.7 exceeds the 10% critical value 7.779 on 4 degrees of freedom, so H0H_{0} is retained and the data does not support the policeman's claim.
  4. D14.714.7 falls below the 10% critical value 15.51 on 8 degrees of freedom, so H0H_{0} is retained and the data does not support the policeman's claim.

Section a, Question 5

[2 marks]Poisson goodness of fit

One hundred electrical components are tested to see how many defects each has. The results are shown in the table.

number of defects0123456≥7\ge 7
number of components112226249530

Calculate the mean number of defects per component.

Answer this when you sit the paper.

[2 marks]Poisson goodness of fit

One hundred electrical components are tested to see how many defects each has. The results are shown in the table.

number of defects0123456≥7\ge 7
number of components112226249530

The mean of this distribution is 2.25. Calculate the expected frequency of components with 0 defects for the Poisson distribution having the same mean, correct to 1 decimal place.

Answer this when you sit the paper.

[2 marks]Poisson goodness of fit

One hundred electrical components are tested to see how many defects each has. The results are shown in the table.

number of defects0123456≥7\ge 7
number of components112226249530

The mean of this distribution is 2.25. Calculate the expected frequency of components with 3 defects for the Poisson distribution having the same mean, correct to 1 decimal place.

Answer this when you sit the paper.

[2 marks]Poisson goodness of fit

One hundred electrical components are tested to see how many defects each has. The results are shown in the table.

number of defects0123456≥7\ge 7
number of components112226249530

The expected Poisson frequencies are 10.5, 23.7, 26.7, 20.0, 11.3, 5.1, 1.9 and 0.8 for 0, 1, 2, 3, 4, 5, 6 and ≥7\ge 7 defects. The classes 5, 6 and ≥7\ge 7 are pooled into a single class ≥5\ge 5, and the mean is estimated from the data.

State the number of degrees of freedom for the goodness of fit test.

Answer this when you sit the paper.

[3 marks]Poisson goodness of fit

One hundred electrical components are tested to see how many defects each has. The results are shown in the table.

number of defects0123456≥7\ge 7
number of components112226249530

The classes 5, 6 and ≥7\ge 7 are pooled into one class ≥5\ge 5, giving observed frequencies 11, 22, 26, 24, 9 and 8 against expected Poisson frequencies 10.5, 23.7, 26.7, 20.0, 11.3 and 7.8.

Calculate the value of the chi-squared test statistic.

Answer this when you sit the paper.

[3 marks]Poisson goodness of fit

One hundred electrical components are tested to see how many defects each has. The results are shown in the table.

number of defects0123456≥7\ge 7
number of components112226249530

A chi-squared goodness of fit test at the 5% level of significance, against the Poisson distribution with the same mean, gives χ2=1.437\chi^{2} = 1.437 on 4 degrees of freedom.

What is the conclusion?

  1. A1.4371.437 is so small that the fit is closer than chance would allow, so H0H_{0} is rejected and the data must have been adjusted.
  2. B1.4371.437 is well below the 5% critical value 9.488 on 4 degrees of freedom, so H0H_{0} is rejected and the data do not come from a Poisson distribution.
  3. C1.4371.437 is well below the 5% critical value 9.488 on 4 degrees of freedom, so H0H_{0} is not rejected and the data are consistent with a Poisson distribution.
  4. D1.4371.437 is well below the 5% critical value 11.070 on 5 degrees of freedom, so H0H_{0} is not rejected and the data are consistent with a Poisson distribution.

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