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ZIMSEC A Level · 9164/4 · J2017

Statistics Paper 4 June 2017

Questions
41
Total marks
94
Syllabus code
9164/4

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Questions
41
Pass mark
25
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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]Geometric distribution

X is a discrete random variable with a geometric distribution, X∼Geo(0.4)X \sim Geo(0.4), counting the number of trials up to and including the first success.

Find P(X≤7)P(X \le 7).

Answer this when you sit the paper.

[3 marks]Geometric distribution

X is a discrete random variable with a geometric distribution, X∼Geo(0.4)X \sim Geo(0.4), counting the number of trials up to and including the first success.

Find P(X>8∣X>3)P(X > 8 \mid X > 3).

Answer this when you sit the paper.

[2 marks]Poisson approximation to the binomial

A survey of 2 000 students at a certain university shows that on average one in every 500 students catches a cold in a week.

Using a suitable approximation, find the probability that exactly one student catches a cold in a week.

Answer this when you sit the paper.

[3 marks]Poisson approximation to the binomial

A survey of 2 000 students at a certain university shows that on average one in every 500 students catches a cold in a week, so the mean number catching a cold in a week is 4.

Using a Poisson model, find the probability that at least three students catch a cold in a month, assuming the month has exactly 28 days.

  1. A0.000016317
  2. B0.2381
  3. C0.7619
  4. D0.9999837
[2 marks]Median of a data set

The marks obtained by 36 advanced level students in a Mathematics test were

59, 53, 74, 55, 90, 57, 88, 68, 59, 67, 82, 62, 61, 77, 74, 86, 60, 83, 92, 58, 60, 72, 57, 96, 56, 67, 73, 78, 66, 79, 51, 60, 54, 67, 80, 63

Find the median mark.

Answer this when you sit the paper.

[2 marks]Quartiles and interquartile range

The marks obtained by 36 advanced level students in a Mathematics test were

59, 53, 74, 55, 90, 57, 88, 68, 59, 67, 82, 62, 61, 77, 74, 86, 60, 83, 92, 58, 60, 72, 57, 96, 56, 67, 73, 78, 66, 79, 51, 60, 54, 67, 80, 63

Find the interquartile range of these marks.

Answer this when you sit the paper.

[2 marks]Stem and leaf diagrams

The marks obtained by 36 advanced level students in a Mathematics test were

59, 53, 74, 55, 90, 57, 88, 68, 59, 67, 82, 62, 61, 77, 74, 86, 60, 83, 92, 58, 60, 72, 57, 96, 56, 67, 73, 78, 66, 79, 51, 60, 54, 67, 80, 63

The marks are grouped into class intervals of width 5 marks, so one class holds the marks from 65 to 69 inclusive. How many of the 36 marks fall in that class?

  1. A3
  2. B4
  3. C5
  4. D6
[2 marks]Continuous random variables

The continuous random variable X has probability density function f(x)=2e−kxf(x) = 2e^{-kx} for x≥0x \ge 0 and f(x)=0f(x) = 0 for x<0x < 0, where k is a positive integer.

In terms of k, what is the value of ∫0∞2e−kx dx\displaystyle\int_{0}^{\infty} 2e^{-kx}\,dx?

  1. A2k2k
  2. B1k\dfrac{1}{k}
  3. C2k\dfrac{2}{k}
  4. Dk2\dfrac{k}{2}
[2 marks]Continuous random variables

The continuous random variable X has probability density function f(x)=2e−2xf(x) = 2e^{-2x} for x≥0x \ge 0 and f(x)=0f(x) = 0 for x<0x < 0.

Which of these is the cumulative distribution function of X?

  1. AF(x)=0F(x) = 0 for x<0x < 0 and F(x)=1−e−2xF(x) = 1 - e^{-2x} for x≥0x \ge 0
  2. BF(x)=0F(x) = 0 for x<0x < 0 and F(x)=e−2x−1F(x) = e^{-2x} - 1 for x≥0x \ge 0
  3. CF(x)=0F(x) = 0 for x<0x < 0 and F(x)=1−2e−2xF(x) = 1 - 2e^{-2x} for x≥0x \ge 0
  4. DF(x)=0F(x) = 0 for x<0x < 0 and F(x)=2−2e−2xF(x) = 2 - 2e^{-2x} for x≥0x \ge 0
[2 marks]Continuous random variables

The continuous random variable X has probability density function f(x)=2e−2xf(x) = 2e^{-2x} for x≥0x \ge 0 and f(x)=0f(x) = 0 for x<0x < 0, so its cumulative distribution function is F(x)=1−e−2xF(x) = 1 - e^{-2x} for x≥0x \ge 0.

Find the exact value of the median of X.

Answer this when you sit the paper.

Section a, Question 3

[2 marks]Normal distribution, finding mean and standard deviation

The random variable X is normally distributed with mean μ\mu and standard deviation σ\sigma, and P(X>65)=0.01P(X > 65) = 0.01.

Write down the value of z for which Φ(z)=0.99\Phi(z) = 0.99, giving your answer correct to 3 decimal places.

Answer this when you sit the paper.

[3 marks]Normal distribution, finding mean and standard deviation
The random variable X is normally distributed with mean μ\mu and variance σ2\sigma^{2}. Given that P(X>65)=0.01P(X > 65) = 0.01 and P(X<20)=0.02P(X < 20) = 0.02, find σ\sigma.

Answer this when you sit the paper.

[2 marks]Normal distribution, finding mean and standard deviation

The random variable X is normally distributed with mean μ\mu and variance σ2\sigma^{2}, with P(X>65)=0.01P(X > 65) = 0.01 and P(X<20)=0.02P(X < 20) = 0.02. Solving the two standardised equations gives σ=10.27\sigma = 10.27.

Find μ\mu.

Answer this when you sit the paper.

[1 marks]Mean of a frequency distribution

The discrete random variable X is distributed as shown in the table below.

X01234
Frequency46442082

Calculate the mean value of X.

Answer this when you sit the paper.

[2 marks]Poisson distribution

A set of 120 observations of a discrete random variable X has mean 2930\dfrac{29}{30}.

A Poisson model with the same mean is fitted to the data. Find the expected frequency of the value X=0X = 0 out of the 120 observations.

Answer this when you sit the paper.

[2 marks]Poisson distribution

A set of 120 observations of a discrete random variable X has mean 2930\dfrac{29}{30}.

A Poisson model with the same mean is fitted to the data. Find the expected frequency of the value X=2X = 2 out of the 120 observations.

Answer this when you sit the paper.

[3 marks]Chi-squared goodness of fit to a Poisson model

A goodness of fit test compares 120 observations of X with a Poisson model of the same mean. The observed and expected frequencies are

X01234 or more
Observed46442082
Expected45.6444.1221.326.872.05

The last two classes are pooled so that no expected frequency is below 5. Calculate the value of the test statistic χ2\chi^{2}.

Answer this when you sit the paper.

[2 marks]Chi-squared goodness of fit to a Poisson model

A goodness of fit test at the 5% level of significance compares 120 observations of X with a Poisson model whose mean was estimated from the same data. After pooling there are four classes, the test statistic is χ2=0.2153\chi^{2} = 0.2153 and the critical value is χ0.052(2)=5.991\chi^{2}_{0.05}(2) = 5.991.

What is the conclusion of the test?

  1. AThe test statistic 0.2153 exceeds the critical value 5.991, so the null hypothesis is not rejected and the data are consistent with a Poisson distribution.
  2. BThe test statistic 0.2153 exceeds the critical value 5.991, so the null hypothesis is rejected and the data do not follow a Poisson distribution.
  3. CThe test statistic 0.2153 is below the critical value 5.991, so the null hypothesis is rejected and the data do not follow a Poisson distribution.
  4. DThe test statistic 0.2153 is below the critical value 5.991, so the null hypothesis is not rejected and the data are consistent with a Poisson distribution.
[2 marks]Chi-squared test of association

An agriculture class applied three fertilisers X, Y and Z to 75 beds of beans and classified the yield per bed as high, medium or low.

YieldXYZ
High12153
Medium888
Low579

A chi-squared test of association is to be carried out. Calculate the expected frequency for a high yield with fertiliser Z.

Answer this when you sit the paper.

[2 marks]Chi-squared test of association

An agriculture class applied three fertilisers X, Y and Z to 75 beds of beans and classified the yield per bed as high, medium or low.

YieldXYZ
High12153
Medium888
Low579

A chi-squared test of association is to be carried out. Calculate the expected frequency for a medium yield with fertiliser Y.

Answer this when you sit the paper.

[3 marks]Chi-squared test of association

An agriculture class applied three fertilisers X, Y and Z to 75 beds of beans and classified the yield per bed as high, medium or low. The observed frequencies are

YieldXYZ
High12153
Medium888
Low579

and the expected frequencies under no association are

YieldXYZ
High10128
Medium89.66.4
Low78.45.6

Calculate the value of the test statistic χ2\chi^{2}.

Answer this when you sit the paper.

[2 marks]Chi-squared test of association

A chi-squared test of association is carried out at the 1% level of significance on a table with 3 rows and 3 columns.

Write down the critical value χ0.012\chi^{2}_{0.01} for this test.

Answer this when you sit the paper.

[2 marks]Chi-squared test of association

A chi-squared test of association between type of fertiliser and yield is carried out at the 1% level of significance on a 3 by 3 table. The test statistic is χ2=7.810\chi^{2} = 7.810 and the critical value is χ0.012(4)=13.277\chi^{2}_{0.01}(4) = 13.277.

What is the conclusion of the test?

  1. A7.810 is less than 13.277, so the null hypothesis is rejected: there is evidence of an association between type of fertiliser and yield.
  2. B7.810 is greater than 13.277, so the null hypothesis is not rejected: there is no evidence of an association between fertiliser and yield.
  3. C7.810 is less than 13.277, so the null hypothesis is not rejected: there is no evidence of an association between fertiliser and yield.
  4. D7.810 is greater than 13.277, so the null hypothesis is rejected: there is evidence of an association between type of fertiliser and yield.
[3 marks]Normal distribution

The random variable R is normally distributed with R∼N(54,36)R \sim N(54, 36).

Find the value of r such that P(R≤r)=0.484P(R \le r) = 0.484.

Answer this when you sit the paper.

[3 marks]Normal distribution

The random variable S is normally distributed with S∼N(48,25)S \sim N(48, 25).

Find the value of s such that P(S≥s)=0.484P(S \ge s) = 0.484.

Answer this when you sit the paper.

[2 marks]Normal distribution, sums and differences

The random variables R and S are independent and normally distributed, with R∼N(54,36)R \sim N(54, 36) and S∼N(48,25)S \sim N(48, 25).

What is the distribution of R−SR - S?

  1. AR−S∼N(6, 61)R - S \sim N(6,\ 61)
  2. BR−S∼N(102, 61)R - S \sim N(102,\ 61)
  3. CR−S∼N(6, 1)R - S \sim N(6,\ 1)
  4. DR−S∼N(6, 11)R - S \sim N(6,\ 11)
[3 marks]Normal distribution, sums and differences

The random variables R and S are independent and normally distributed, with R∼N(54,36)R \sim N(54, 36) and S∼N(48,25)S \sim N(48, 25).

Find P(R≥S)P(R \ge S).

Answer this when you sit the paper.

[3 marks]Normal distribution, sums and differences

The random variable R is normally distributed with R∼N(54,36)R \sim N(54, 36). Six independent observations of R are taken.

Find the probability that the sum of the six observations is less than 300.

Answer this when you sit the paper.

Section a, Question 4

[2 marks]Coded summary statistics

The marks x obtained by a random sample of n students in a test are summarised by ∑(x−25)=144\sum(x - 25) = 144, and the sample mean is xˉ=28.6\bar{x} = 28.6.

Find the value of n.

Answer this when you sit the paper.

[2 marks]Coded summary statistics

The marks x obtained by a random sample of 40 students in a test have mean xˉ=28.6\bar{x} = 28.6.

Find ∑x\sum x.

Answer this when you sit the paper.

[2 marks]Coded summary statistics

For a sample of 40 marks with ∑x=1144\sum x = 1144, the coded sum of squares is ∑(x−25)2=3650\sum(x - 25)^{2} = 3650.

Find ∑x2\sum x^{2}.

Answer this when you sit the paper.

[2 marks]Unbiased estimates

A random sample of 40 marks has ∑x=1144\sum x = 1144 and ∑x2=35 850\sum x^{2} = 35\,850.

Calculate the unbiased estimate of the population variance.

Answer this when you sit the paper.

[3 marks]Confidence intervals

A random sample of 40 marks has mean 28.6 and unbiased estimate of the population variance 80.3.

Which of these is the 99% confidence interval for the population mean?

  1. A(23.43, 33.77)
  2. B(24.95, 32.25)
  3. C(25.82, 31.38)
  4. D(26.27, 30.93)
[3 marks]Hypothesis testing for a mean

A random sample of 40 marks from a normal population has mean 28.6 and unbiased estimate of the population variance 80.3.

The hypothesis μ=30\mu = 30 is tested against the alternative μ<30\mu < 30. Calculate the value of the test statistic z.

Answer this when you sit the paper.

[2 marks]Hypothesis testing for a mean

A one-tailed test at the 5% level of significance of μ=30\mu = 30 against μ<30\mu < 30 gives a test statistic of z=−0.988z = -0.988. The critical value is z=−1.645z = -1.645.

What is the conclusion of the test?

  1. A-0.988 is greater than -1.645, so the null hypothesis is rejected: there is evidence that the mean mark is less than 30.
  2. B-0.988 is greater than -1.645, so the null hypothesis is not rejected: there is no evidence that the mean mark is below 30.
  3. C-0.988 is less than -1.645, so the null hypothesis is not rejected: there is no evidence that the mean mark is below 30.
  4. D-0.988 is less than -1.645, so the null hypothesis is rejected: there is evidence that the mean mark is less than 30.
[1 marks]Scatter diagrams, regression and correlation

Ten candidates obtained the following marks, where X is the paper 1 mark and Y is the paper 2 mark.

X86937366889680709563
Y71766152759471608555

Find the mean paper 1 mark, xˉ\bar{x}.

Answer this when you sit the paper.

[3 marks]Scatter diagrams, regression and correlation

Ten candidates obtained the following marks, where X is the paper 1 mark and Y is the paper 2 mark.

X86937366889680709563
Y71766152759471608555

For these data ∑x=810\sum x = 810, ∑y=700\sum y = 700, ∑x2=66 984\sum x^{2} = 66\,984 and ∑xy=58 103\sum xy = 58\,103.

Find the gradient m of the regression line of Y on X, in the form y=mx+cy = mx + c.

Answer this when you sit the paper.

[3 marks]Scatter diagrams, regression and correlation

For ten candidates, the paper 1 marks have mean xˉ=81\bar{x} = 81 and the paper 2 marks have mean yˉ=70\bar{y} = 70. The regression line of Y on X has gradient m=1.021m = 1.021.

Find the intercept c in y=mx+cy = mx + c.

Answer this when you sit the paper.

[2 marks]Scatter diagrams, regression and correlation

For ten candidates the regression line of the paper 2 mark Y on the paper 1 mark X is y=1.021x−12.7y = 1.021x - 12.7.

Estimate the paper 2 mark for a candidate who has a paper 1 mark of 75, giving your answer to the nearest whole mark.

Answer this when you sit the paper.

[3 marks]Scatter diagrams, regression and correlation

For ten candidates' paper 1 marks X and paper 2 marks Y,

n∑xy−∑x∑y=14 030n\sum xy - \sum x \sum y = 14\,030, n∑x2−(∑x)2=13 740n\sum x^{2} - \left(\sum x\right)^{2} = 13\,740 and n∑y2−(∑y)2=15 940n\sum y^{2} - \left(\sum y\right)^{2} = 15\,940.

Calculate the product moment correlation coefficient.

Answer this when you sit the paper.

[2 marks]Scatter diagrams, regression and correlation

For ten candidates, the product moment correlation coefficient between the paper 1 mark and the paper 2 mark is r=0.948r = 0.948.

Which comment on this value is correct?

  1. AThe value is close to zero, so there is almost no linear relationship between the paper 1 and paper 2 marks.
  2. BThe value is close to 1, so the paper 1 mark is what caused the paper 2 mark for each of the ten candidates.
  3. CThe value is close to minus 1, so there is a very strong negative linear relationship between the two marks.
  4. DThe value is close to 1, so there is a very strong positive linear relationship between the paper 1 and paper 2 marks.

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