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ZIMSEC A Level · 9164/4 · J2008

Statistics Paper 4 June 2008

Questions
25
Total marks
38
Syllabus code
9164/4

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Questions
25
Pass mark
15
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Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Question 101

[1 marks]confidence intervals
The diameters of 25 steel rods have a sample mean of 0.980 cm and a standard deviation of 0.015 cm. Assuming the diameters are normally distributed with this standard deviation, find the 99% confidence interval for the population mean diameter.
  1. A(0.9730, 0.9870)(0.9730,\ 0.9870) cm
  2. B(0.9741, 0.9859)(0.9741,\ 0.9859) cm
  3. C(0.9723, 0.9877)(0.9723,\ 0.9877) cm
  4. D(0.9414, 1.0186)(0.9414,\ 1.0186) cm

Question 201

[1 marks]statistics - stem and leaf
The masses (g) of 24 sweets, read off a stem-and-leaf diagram and placed in ascending order, are: 0.72, 0.73, 0.79, 0.80, 0.88, 0.91, 0.91, 0.94, 0.98, 0.99, 1.01, 1.03, 1.06, 1.08, 1.13, 1.13, 1.13, 1.19, 1.21, 1.22, 1.33, 1.39, 1.44, 1.45. Find the median mass.
  1. A1.031.03 g
  2. B1.0451.045 g
  3. C1.061.06 g
  4. D1.071.07 g

Question 202

[1 marks]statistics - stem and leaf
Using the same 24 sweet masses (g) 0.72, 0.73, 0.79, 0.80, 0.88, 0.91, 0.91, 0.94, 0.98, 0.99, 1.01, 1.03, 1.06, 1.08, 1.13, 1.13, 1.13, 1.19, 1.21, 1.22, 1.33, 1.39, 1.44, 1.45, find the mode.
  1. A1.211.21 g
  2. B0.910.91 g
  3. C1.031.03 g
  4. D1.131.13 g

Question 203

[1 marks]statistics - stem and leaf
A sweet with mass greater than 1.2 g is classified as large. From the 24 masses (g) 0.72, 0.73, 0.79, 0.80, 0.88, 0.91, 0.91, 0.94, 0.98, 0.99, 1.01, 1.03, 1.06, 1.08, 1.13, 1.13, 1.13, 1.19, 1.21, 1.22, 1.33, 1.39, 1.44, 1.45, calculate the mean mass of the large sweets.
  1. A1.341.34 g
  2. B1.371.37 g
  3. C1.321.32 g
  4. D0.340.34 g

Question 301

[1 marks]probability
A fair die is tossed three times. Find the probability that exactly one six is obtained.
  1. A25216\frac{25}{216}
  2. B2572\frac{25}{72}
  3. C536\frac{5}{36}
  4. D16\frac{1}{6}

Question 302

[1 marks]probability
A fair die is tossed three times. Find the probability that the first score is even, the second score is odd, and the third score is either a 1 or a 2.
  1. A18\frac{1}{8}
  2. B16\frac{1}{6}
  3. C136\frac{1}{36}
  4. D112\frac{1}{12}

Question 401

[1 marks]hypothesis testing
A dairy farmer claims his milk bottles contain exactly one litre. A random sample of 20 bottles has mean 0.980 litres and standard deviation 0.070 litres. Calculate the test statistic for testing H0:μ=1H_0:\mu=1 against H1:μ≠1H_1:\mu\neq1.
  1. At≈−5.71t \approx -5.71
  2. Bt≈1.28t \approx 1.28
  3. Ct≈−0.29t \approx -0.29
  4. Dt≈−1.28t \approx -1.28

Question 402

[1 marks]hypothesis testing
For the milk-bottle test (n=20n=20, sample mean 0.980 litres, sample standard deviation 0.070 litres, testing H0:μ=1H_0:\mu=1 against H1:μ≠1H_1:\mu\neq1 at the 5% significance level, with critical value t19,0.025=2.093t_{19,0.025}=2.093 and calculated test statistic t≈−1.28t\approx-1.28), what is the correct conclusion?
  1. AThe test cannot be completed because 0.980 litres is numerically less than 1 litre.
  2. BFail to reject H0H_0; there is insufficient evidence at the 5% level that the mean content differs from 1 litre.
  3. CReject H0H_0; there is significant evidence the mean content differs from 1 litre.
  4. DFail to reject H0H_0, but only because the degrees of freedom should be taken as 20, not 19.

Question 501

[1 marks]normal distribution
A pie-making machine produces pie masses that are normally distributed with standard deviation 0.8 g. Legal requirements specify that at most 0.3% of pies may have mass below 80 g, and this limit is met exactly, so P(X<80)=0.003P(X<80)=0.003. Find the mean mass μ\mu.
  1. Aμ≈84.40\mu \approx 84.40 g
  2. Bμ≈82.20\mu \approx 82.20 g
  3. Cμ≈77.80\mu \approx 77.80 g
  4. Dμ≈80.80\mu \approx 80.80 g

Question 502

[1 marks]normal distribution
Continuing the pie scenario: masses are normally distributed with mean 82.20 g and standard deviation 0.8 g, the weekly output is 500 000 pies, and pies with mass over 83 g each cost an extra \$10 to package. Find the firm's weekly extra-packaging cost.
  1. A$793 500\$793\,500
  2. B$500\$500
  3. C$4 206 500\$4\,206\,500
  4. D$79 350\$79\,350

Question 601

[1 marks]discrete random variables
A discrete random variable XX takes the values 0, 1 and 2 with probabilities P0P_0, P1P_1 and P2P_2 respectively. Given E(X)=43E(X)=\frac43 and Var(X)=59\mathrm{Var}(X)=\frac59, find P0P_0, P1P_1 and P2P_2.
  1. AP0=16, P1=13, P2=12P_0=\frac16,\ P_1=\frac13,\ P_2=\frac12
  2. BP0=0, P1=23, P2=13P_0=0,\ P_1=\frac23,\ P_2=\frac13
  3. CP0=16, P1=12, P2=13P_0=\frac16,\ P_1=\frac12,\ P_2=\frac13
  4. DP0=0, P1=13, P2=23P_0=0,\ P_1=\frac13,\ P_2=\frac23

Question 701

[1 marks]probability distributions
A random variable YY has distribution P(Y=4)=0.14P(Y=4)=0.14, P(Y=8)=0.17P(Y=8)=0.17, P(Y=9)=0.19P(Y=9)=0.19, P(Y=k)=0.29P(Y=k)=0.29, P(Y=16)=pP(Y=16)=p, with E(Y)=10.18E(Y)=10.18. Find pp and kk.
  1. Ap=0.19, k=11p=0.19,\ k=11
  2. Bp=0.21, k=10p=0.21,\ k=10
  3. Cp=0.21, k=11p=0.21,\ k=11
  4. Dp=0.21, k=12p=0.21,\ k=12

Question 702

[1 marks]probability distributions
For the random variable YY above, with P(Y=4)=0.14P(Y=4)=0.14, P(Y=8)=0.17P(Y=8)=0.17, P(Y=9)=0.19P(Y=9)=0.19, P(Y=11)=0.29P(Y=11)=0.29, P(Y=16)=0.21P(Y=16)=0.21 and E(Y)=10.18E(Y)=10.18, find Var(Y)\mathrm{Var}(Y).
  1. A13.7313.73
  2. B97.0097.00
  3. C107.18107.18
  4. D117.36117.36

Question 703

[1 marks]probability distributions
A random sample of 31 observations of YY is taken, where E(Y)=10.18E(Y)=10.18 and Var(Y)=13.73\mathrm{Var}(Y)=13.73. Find P(Yˉ>9.78)P(\bar Y>9.78), correct to 2 decimal places.
  1. A0.270.27
  2. B0.500.50
  3. C0.600.60
  4. D0.730.73

Question 801

[1 marks]Poisson distribution
An airline experiences delays at an average rate of 1 per two weeks. Find the probability that at least 2 delays occur in a particular 3-week period.
  1. A0.26420.2642
  2. B0.44220.4422
  3. C0.55780.5578
  4. D0.77690.7769

Question 802

[1 marks]Poisson distribution
The airline's financial year consists of 17 independent three-week periods. Using P(at least 2 delays in a 3-week period)=0.4422P(\text{at least 2 delays in a 3-week period})=0.4422, let WW be the number of these 17 periods during which at least 2 delays occur. Find P(W≤2)P(W\le2).
  1. A0.00490.0049
  2. B0.08100.0810
  3. C0.44220.4422
  4. D0.99510.9951

Question 803

[1 marks]Poisson distribution
Delays occur at a rate of 1 per two weeks, so in an nn-week period the expected number of delays is λ=n/2\lambda=n/2. Given that the probability of at least one delay occurring in a period of nn weeks is greater than 0.875, find the least possible integer value of nn.
  1. An=3n=3
  2. Bn=9n=9
  3. Cn=4n=4
  4. Dn=5n=5

Question 901

[1 marks]chi-squared test
In a job-satisfaction survey of 310 employees classified by satisfaction level (High/Medium/Low) and salary band, the row totals are 50 (High), 200 (Medium), 60 (Low) and the column totals are 160 (under \$10M), 100 (\$10-20M), 50 (over \$20M), against observed counts of 20, 20, 10 (High row), 100, 65, 35 (Medium row) and 40, 15, 5 (Low row). Calculate the chi-squared test statistic ∑(O−E)2E\sum\frac{(O-E)^2}{E}.
  1. A4.464.46
  2. B6.176.17
  3. C8.918.91
  4. D17.8217.82

Question 902

[1 marks]chi-squared test
Using the calculated chi-squared statistic χ2≈8.91\chi^2\approx8.91 for the 3×33\times3 job-satisfaction table above, with (3−1)(3−1)=4(3-1)(3-1)=4 degrees of freedom and a 5% critical value of χ4,0.052=9.488\chi^2_{4,0.05}=9.488, what conclusion should be drawn about salary and job satisfaction?
  1. AFail to reject H0H_0: there is insufficient evidence at the 5% level that salary and job satisfaction are associated.
  2. BReject H0H_0: salary and job satisfaction are significantly associated at the 5% level.
  3. CFail to reject H0H_0, but only because 6 degrees of freedom should have been used instead of 4.
  4. DThe test cannot be completed because one expected frequency is below 5.

Question 1001

[1 marks]continuous random variables
A continuous random variable XX has probability density function f(x)=x4f(x)=\frac{x}{4} for 1≤x≤31\le x\le3 (zero otherwise). Find Var(X)\mathrm{Var}(X).
  1. A55
  2. B176\frac{17}{6}
  3. C136\frac{13}{6}
  4. D1136\frac{11}{36}

Question 1002

[1 marks]continuous random variables
Independent random variables XX and YY have E(X)=136E(X)=\frac{13}{6} and E(Y)=3E(Y)=3, where YY has pdf g(y)=12g(y)=\frac12 for 2≤y≤42\le y\le4. Find E(4Y−3X)E(4Y-3X).
  1. A−5.50-5.50
  2. B−0.33-0.33
  3. C0.830.83
  4. D5.505.50

Question 1003

[1 marks]continuous random variables
With Var(X)=1136\mathrm{Var}(X)=\frac{11}{36} and Var(Y)=13\mathrm{Var}(Y)=\frac13, and X,YX,Y independent, find Var(4Y−3X)\mathrm{Var}(4Y-3X) to 2 decimal places.
  1. A2.252.25
  2. B2.582.58
  3. C7.897.89
  4. D8.088.08

Question 1101

[1 marks]regression and correlation
For a sample of 6 workshop participants, dd is the distance driven (km) and tt is the time taken (min), with ∑(d−300)=394\sum(d-300)=394, ∑(d−300)2=123648\sum(d-300)^2=123648, ∑(t−200)=570\sum(t-200)=570, ∑(t−200)2=103500\sum(t-200)^2=103500, ∑(d−300)(t−200)=103930\sum(d-300)(t-200)=103930. Find the equation of the estimated regression line of tt on dd.
  1. At=365.67+0.680dt=365.67+0.680d
  2. Bt=−12.36+0.840dt=-12.36+0.840d
  3. Ct=46.35+0.680dt=46.35+0.680d
  4. Dt=46.35+1.470dt=46.35+1.470d

Question 1102

[1 marks]regression and correlation
Using the regression line t=46.35+0.680dt=46.35+0.680d obtained above, estimate the time taken by a participant who drove 350 km.
  1. A318318 minutes
  2. B284284 minutes
  3. C4646 minutes
  4. D238238 minutes

Question 1103

[1 marks]regression and correlation
Using Sdd=97777.33S_{dd}=97777.33, Stt=49350S_{tt}=49350 and Sdt=66500S_{dt}=66500 from the workshop data, find the product-moment correlation coefficient between dd and tt, and comment on it.
  1. Ar≈0.92r\approx0.92, reported as the correlation coefficient.
  2. Br≈0.68r\approx0.68, indicating a moderate positive correlation.
  3. Cr≈0.96r\approx0.96, indicating a strong positive linear correlation between distance and time.
  4. Dr≈0.96r\approx0.96, indicating a strong negative linear correlation between distance and time.

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