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Normal distribution and further hypothesis testingAQA A-Level Maths: Topic test

20 questions, 54 marks

AQA A-Level Maths

Normal distribution and further hypothesis testing topic test

Total 54 marks

Name

Class

Date

  1. 1
    The wingspan WW cm of adult dragonflies of a certain species is modelled by the Normal distribution W∼N(8.4,0.62)W\sim N(8.4,0.6^2).
    (a)
    What is P(W<7.8)P(W<7.8)?
    [1 mark]
    • A0.04780.0478
    • B0.84130.8413
    • C0.15870.1587
    • D0.02280.0228
    (b)
    At which values of WW does the graph of the probability density function of WW have points of inflection?
    [1 mark]
    • A7.87.8 and 9.09.0
    • B7.27.2 and 9.69.6
    • C8.048.04 and 8.768.76
    • D8.48.4 only
    (c)
    Find the probability that the wingspan of a randomly chosen dragonfly is within 1.21.2 cm of the mean.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    A factory produces components, each of which is defective with probability 0.030.03. An inspector takes a random sample of 88 components from a large batch and counts the number XX that are defective.
    (a)
    Which distribution is the most appropriate model for XX?
    [1 mark]
    • AX∼N(8,0.03)X\sim N(8,0.03)
    • BX∼B(8,0.03)X\sim B(8,0.03)
    • CX∼B(8,0.97)X\sim B(8,0.97)
    • DX∼N(0.24,0.2328)X\sim N(0.24,0.2328)
    (b)
    A student suggests approximating the distribution of XX by a Normal distribution. Which statement gives the best reason why this is unsuitable?
    [1 mark]
    • AA binomial distribution can never be approximated by a Normal distribution.
    • BThe probability of a defect, 0.030.03, is less than 0.50.5.
    • CXX takes only whole-number values, so no continuous model can ever be used.
    • DThe mean of XX is only 0.240.24, so its distribution is strongly skewed and a Normal model would give a large probability of negative values.
    (c)
    State two conditions that must hold for B(8,0.03)B(8,0.03) to be a valid model for XX.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    The resting heart rate of adult members of a running club has historically been Normally distributed with mean 6262 beats per minute and standard deviation 55. After a new training programme, the coach believes that the mean resting heart rate has fallen. A random sample of 1616 members is taken and the sample mean is 59.559.5 beats per minute. Assume that the standard deviation is unchanged.
    (a)
    State suitable hypotheses for the coach's test, defining any symbol you use, and state the distribution of the sample mean Xˉ\bar X if the null hypothesis is true.
    [3 marks]
    (b)
    Carry out the test at the 5%5\% significance level, stating your conclusion in context.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    A machine cuts planks of timber. The length LL cm of a plank is modelled by L∼N(200,σ2)L\sim N(200,\sigma^2), where σ\sigma is unknown. It is found that 5%5\% of planks are longer than 202.5202.5 cm.
    (a)
    (i) Show that σ=1.52\sigma=1.52, correct to three significant figures. (ii) Find the probability that a randomly chosen plank is shorter than 198198 cm.
    [6 marks]
    (b)
    After the machine is adjusted, a random sample of 2525 planks has mean length 199.4199.4 cm. Assuming that the standard deviation is still 1.521.52 cm, test at the 5%5\% level whether the mean length has changed. State one assumption you need to make about the lengths of the planks.
    [6 marks]

    Total for question 4: 12 marks

  5. 5
    The score XX on a reaction test is modelled by X∼N(50,16)X\sim N(50,16).
    (a)
    What is P(X>54)P(X>54)?
    [1 mark]
    • A0.15870.1587
    • B0.84130.8413
    • C0.40130.4013
    • D0.02280.0228
    (b)
    Find the value kk such that P(X<k)=0.9P(X<k)=0.9.
    [1 mark]
    • A51.351.3
    • B53.453.4
    • C55.155.1
    • D70.570.5
    (c)
    Five different people take the test independently. Find the probability that exactly two of them score more than 5454.
    [2 marks]

    Total for question 5: 4 marks

  6. 6
    A zoologist measures the body length and the mass of each of 3030 randomly chosen adult otters. The product moment correlation coefficient between body length and mass is r=0.52r=0.52. The critical value of the product moment correlation coefficient for a two-tailed test at the 5%5\% significance level with n=30n=30 is 0.36100.3610.
    (a)
    The zoologist tests whether there is any correlation between body length and mass in the population. Which pair of hypotheses should be used?
    [1 mark]
    • AH0:r=0H_0:r=0, H1:r≠0H_1:r\ne0
    • BH0:ρ≠0H_0:\rho\ne0, H1:ρ=0H_1:\rho=0
    • CH0:ρ=0H_0:\rho=0, H1:ρ>0H_1:\rho>0
    • DH0:ρ=0H_0:\rho=0, H1:ρ≠0H_1:\rho\ne0
    (b)
    Which conclusion is correct?
    [1 mark]
    • ASince 0.52<10.52<1, there is no evidence of correlation.
    • BSince 0.52>0.36100.52>0.3610, reject H0H_0: there is sufficient evidence at the 5%5\% level of correlation between body length and mass.
    • CSince 0.52>0.36100.52>0.3610, the test proves that greater body length causes greater mass.
    • DSince 0.52>0.36100.52>0.3610, accept H0H_0: there is no correlation between body length and mass.
    (c)
    Explain why the result of the test does not show that greater body length causes greater mass.
    [2 marks]

    Total for question 6: 4 marks

  7. 7
    A dealer investigates whether the age of a used car and its advertised price are negatively correlated. A product moment correlation coefficient is calculated for each random sample of cars, and a one-tailed test is carried out at the 5%5\% significance level.
    (a)
    A random sample of 1515 cars gives r=−0.38r=-0.38. The critical value for a one-tailed test at the 5%5\% level with n=15n=15 is 0.44090.4409. State the hypotheses and the conclusion of the test, in context.
    [3 marks]
    (b)
    A second random sample of 3030 cars also gives r=−0.38r=-0.38. The critical value for a one-tailed test at the 5%5\% level with n=30n=30 is 0.30610.3061. (i) Carry out the test using this sample. (ii) Explain why the conclusions of the two tests differ even though rr is the same.
    [4 marks]

    Total for question 7: 7 marks

  8. 8
    The reaction time RR milliseconds of sprinters in a large population is modelled by R∼N(180,202)R\sim N(180,20^2).
    (a)
    (i) Find P(R<150)P(R<150). (ii) Find the value of rr such that 2.5%2.5\% of sprinters have a reaction time greater than rr. (iii) Four sprinters are chosen at random. Find the probability that at least one of them has a reaction time below 150150 milliseconds.
    [6 marks]
    (b)
    A coach measures RR and the 100100 m time for each of 2525 randomly chosen sprinters and finds a product moment correlation coefficient of 0.470.47. The critical value for a two-tailed test at the 5%5\% level with n=25n=25 is 0.39610.3961. (i) Test for correlation between RR and 100100 m time, stating your conclusion in context. (ii) Reaction times cannot be below 100100 milliseconds in practice. Show by calculation that the Normal model is still reasonable, despite this.
    [6 marks]

    Total for question 8: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).