Statistics Paper 4 June 2012

X is a continuous random variable with probability density function

$$f(x)=\begin{cases}2-2x & 0 \le x \le 1\\ 0 & \text{otherwise.}\end{cases}$$

Which of these is the cumulative distribution function F(x)?

X is a continuous random variable with probability density function $f(x)=2-2x$ for $0 \le x \le 1$ and $f(x)=0$ otherwise, so its cumulative distribution function is $F(x)=2x-x^{2}$ on $0 \le x \le 1$.

Find $P\left(X \le \dfrac{1}{3}\right)$, giving your answer as an exact fraction.

X is a continuous random variable with probability density function $f(x)=2-2x$ for $0 \le x \le 1$ and $f(x)=0$ otherwise, so its cumulative distribution function is $F(x)=2x-x^{2}$ on $0 \le x \le 1$.

Find the value of p such that $P(X < p) = \dfrac{1}{5}$.

In a certain court there are only two verdicts, "convicted" or "discharged". Of all the cases tried by this court, 80% of the verdicts were convictions. When the verdict is "convicted" the probability that the accused person is innocent is 0.07, and when the verdict is "discharged" the probability that the accused person is innocent is 0.4.

Find the probability that a person tried by this court is innocent.

In a certain court there are only two verdicts, "convicted" or "discharged". Of all the cases tried by this court, 80% of the verdicts were convictions. When the verdict is "convicted" the probability that the accused person is innocent is 0.07, and when the verdict is "discharged" the probability that the accused person is innocent is 0.4.

Find the conditional probability that an innocent person tried by this court is convicted.

The Biology marks of a group of 24 students were

36, 45, 40, 60, 71, 66, 53, 42, 35, 54, 35, 43, 72, 37, 39, 34, 49, 43, 75, 58, 67, 59, 36, 67

Find the median mark.

The Integrated Science marks of a group of 24 students were

88, 89, 30, 34, 48, 49, 59, 65, 67, 78, 41, 70, 54, 66, 39, 49, 37, 59, 45, 63, 52, 75, 38, 38

Find the median mark.

The Biology marks of a group of 24 students were

36, 45, 40, 60, 71, 66, 53, 42, 35, 54, 35, 43, 72, 37, 39, 34, 49, 43, 75, 58, 67, 59, 36, 67

Find the interquartile range of these marks.

For a group of 24 students, the Biology marks have median 47, lower quartile 38 and upper quartile 63. The Integrated Science marks for the same students have median 53, lower quartile 40 and upper quartile 66.5.

Which statement best compares the performance of the students in the two subjects?

The table shows the attitude of 300 parents in three parts of the country towards the introduction of incentives for teachers.

A chi-squared test of association is to be carried out. Calculate the expected frequency for parents in the NORTH who LIKE the introduction of incentives.

The table shows the attitude of 300 parents in three parts of the country towards the introduction of incentives for teachers.

Calculate the value of the chi-squared test statistic for a test of association between area of residence and attitude.

A chi-squared test of association is carried out at the 5 % level on a 3 by 3 table classifying 300 parents by area of residence (NORTH, MIDLANDS, SOUTH) and by attitude (LIKE, RESERVED, DISLIKE). The calculated test statistic is 21.75.

What are the degrees of freedom and the correct conclusion?

The number of patients admitted into a clinic per day was recorded over a period of 30 days.

Calculate the mean number of patients admitted per day, correct to 2 decimal places.

The number of patients admitted into a clinic per day has mean 1.63 and is modelled by a Poisson distribution.

Calculate the probability that the clinic admits exactly 2 patients on a particular day.

The number of patients admitted into a clinic per day has mean 1.63 and is modelled by a Poisson distribution. Admissions on different days are independent.

Calculate the probability that the clinic admits at least 3 patients on each of two consecutive days.

The number of patients admitted into a clinic per day has mean 1.63 and is modelled by a Poisson distribution. The clinic has only four beds available, so patients are turned away when more than four need to be admitted.

Calculate the probability that the clinic will turn away some patients who need to be admitted on a particular day.

A vaccine is applied to 50 samples each of 5 monkeys and the number of living monkeys in each sample was counted after one year.

A binomial distribution $B(5,\ p)$ is to be fitted to these data. Estimate the value of p.

A vaccine is applied to 50 samples each of 5 monkeys. A binomial distribution $B(5,\ 0.212)$ is fitted to the number of living monkeys in a sample.

Calculate the expected frequency, out of the 50 samples, of samples containing exactly 1 living monkey.

A vaccine is applied to 50 samples each of 5 monkeys, giving observed frequencies 17, 20, 9, 2, 1, 1 for 0, 1, 2, 3, 4 and 5 living monkeys. A binomial distribution $B(5,\ 0.212)$ gives expected frequencies 15.19, 20.44, 11.00, 2.96, 0.40 and 0.02.

After pooling every class whose expected frequency is below 5 into a single class "3 or more", calculate the value of the chi-squared goodness of fit statistic.

A binomial goodness of fit test is carried out at the 5 % level on 50 samples of 5 monkeys. After pooling, four classes are compared and the parameter p was estimated from the data. The calculated chi-squared statistic is 0.70.

What are the degrees of freedom and the correct conclusion?

A Form 5 class has 50 students and 5 are to be chosen to attend a seminar. The teacher gives the students tickets numbered 0 to 49. The students holding tickets 0 to 9 put their tickets in a hat and the teacher draws one at random, noting its digit. He then forms his sample from the students whose ticket numbers end in that digit.

Which statement about this method is correct?

A machine should be set up to cut planks 5.00 m long. A random sample of 10 planks cut by the machine had lengths, in metres,

4.94, 4.93, 5.00, 4.76, 5.00, 4.73, 4.63, 5.01, 4.65, 5.03

Calculate the mean length of the sample.

A machine should be set up to cut planks 5.00 m long. A random sample of 10 planks cut by the machine had lengths, in metres,

4.94, 4.93, 5.00, 4.76, 5.00, 4.73, 4.63, 5.01, 4.65, 5.03

The mean of this sample is 4.868 m. Calculate the unbiased estimate of the population standard deviation.

A machine should be set up to cut planks 5.00 m long. A random sample of 10 planks cut by the machine had mean length 4.868 m with unbiased estimate of standard deviation 0.1582 m.

Calculate the value of the t test statistic for testing whether the population mean length is 5.00 m.

A machine should be set up to cut planks 5.00 m long. A sample of 10 planks gives a t test statistic of $-2.639$ for the hypotheses $H_{0}: \mu = 5.00$ against $H_{1}: \mu \ne 5.00$. The test is carried out at the 5 % level and the two-tailed critical values on 9 degrees of freedom are $\pm 2.262$.

What is the correct conclusion?

The weights of broiler chickens are normally distributed with mean 2 kg and standard deviation 0.3 kg. The weights of layers chickens are normally distributed with mean 1.5 kg and standard deviation 0.5 kg.

Determine, correct to 2 significant figures, the probability that a broiler chicken weighs less than 1.8 kg.

The weights of broiler chickens are normally distributed with mean 2 kg and standard deviation 0.3 kg. The weights of layers chickens are normally distributed with mean 1.5 kg and standard deviation 0.5 kg.

Determine, correct to 2 significant figures, the probability that a random sample of 4 broiler chickens weigh more than 8.2 kg in total.

The weights of broiler chickens are normally distributed with mean 2 kg and standard deviation 0.3 kg. The weights of layers chickens are normally distributed with mean 1.5 kg and standard deviation 0.5 kg.

Determine, correct to 2 significant figures, the probability that a layers chicken weighs less than a broiler chicken.

The weights of broiler chickens are normally distributed with mean 2 kg and standard deviation 0.3 kg. The weights of layers chickens are normally distributed with mean 1.5 kg and standard deviation 0.5 kg.

Determine, correct to 2 significant figures, the probability that a random sample of 8 layers chickens weigh more in total than a random sample of 6 broiler chickens.

The marks obtained by 10 candidates in two 'O'-level mathematics papers were

Given that $\sum x = 566$ and $\sum x^{2}=35\,510$, calculate $S_{xx}$.

The marks obtained by 10 candidates in two 'O'-level mathematics papers were

Given that $\sum x = 566$, $\sum y = 474$ and $\sum xy = 30\,792$, calculate $S_{xy}$.

For 10 candidates who sat two 'O'-level mathematics papers, $\bar{x}=56.6$, $\bar{y}=47.4$, $S_{xx}=3474.4$ and $S_{xy}=3963.6$, where x is the Paper 1 mark and y is the Paper 2 mark.

Using the appropriate regression line, estimate the Paper 2 mark of a candidate who scored 65 % in Paper 1. Give your answer to the nearest whole mark.

For 10 candidates who sat two 'O'-level mathematics papers, $\bar{x}=56.6$, $\bar{y}=47.4$, $S_{yy}=5378.4$ and $S_{xy}=3963.6$, where x is the Paper 1 mark and y is the Paper 2 mark.

Using the appropriate regression line, estimate the Paper 1 mark of a candidate who scored 50 % in Paper 2. Give your answer to the nearest whole mark.

The marks of 10 candidates in two 'O'-level mathematics papers give $S_{xx}=3474.4$, $S_{yy}=5378.4$ and $S_{xy}=3963.6$, where x is the Paper 1 mark and y is the Paper 2 mark.

What does a scatter diagram of these data show?

A random sample of 250 candidates in an Olympiad Examination gave $\sum x = 11\,872$ and $\sum x^{2} = 646\,193$, where x is a candidate's mark.

Calculate the unbiased estimate of the population mean mark.

A random sample of 250 candidates in an Olympiad Examination gave $\sum x = 11\,872$ and $\sum x^{2} = 646\,193$, where x is a candidate's mark.

Calculate the unbiased estimate of the population variance.

A random sample of 250 candidates in an Olympiad Examination has mean mark 47.488 and unbiased estimate of standard deviation 18.19.

Calculate the upper limit of a 90 % confidence interval for the population mean mark.

A random sample of 250 candidates in an Olympiad Examination has mean mark 47.488 and standard error of the mean 1.1506.

Calculate the value of the z test statistic for testing $H_{0}: \mu = 49$ against $H_{1}: \mu < 49$.

A one-tailed test of $H_{0}: \mu = 49$ against $H_{1}: \mu < 49$ on a large sample gives a z test statistic of $-1.314$. At the 10 % level the critical value is $-1.282$, and $P(Z < -1.314) = 0.0944$.

For which significance levels $\alpha$ is the null hypothesis rejected?