A.6 Answer: TW 6 tutorial

Answers for Sect. 6.2

Answers embedded. Some answers:

1. In order from left to right: \(70\); \(85\); \(100\); \(115\); \(130\). 3. \(30\) points above the mean. 4. Two standard deviations above the mean. 5. About \(2.5\)%.

A person with an IQ of \(85\) has an IQ that is 15 points below the mean, equivalent to 1 standard deviation(s) below the mean IQ. Using the 68--\(95\)--\(99.7\) rule, the proportion of the population with an IQ less than \(85\) is about 32%. In addition, the proportion of the population with an IQ above \(85\) is about 68%.

Answers for Sect. 6.3

  1. Answers vary.
  2. Answers vary.
  3. Answers vary.
  4. Answers vary. You probably cannot be very accurate using the \(68\)--\(95\)--\(99.7\) rule.
  5. Answers vary.
  6. Two standard deviations from the mean is \(2\times 6.7 = 13.4\), so \(95\)% of females aged \(18\) and over have a measured height between \(161.4 \pm13.4\) approximately, or from \(148.0\)cm to \(174.8\)cm.
  7. As follows:
    • \(z = (171 - 161.4)/6.7 = 1.43\), so the probability is \(0.9236\) or about \(92\)%.
    • So the odds are \(92.36/(100 - 92.36) = 12.1\).
    • The answer is just \(1 - 0.9236 = 0.0764\) or about \(7.6\)%.
    • The probability of over 171cm is \(7.64\)%.
    • So the odds are \(7.64/(100 - 7.64)= 0.083\).
    • \(z\)-scores are \(1.28\) and \(2.78\), so the answer is \(0.9973 - 0.8997 = 0.0976\), or about \(9.8\)%.
    • So the odds are \(9.8/(100 - 9.8) = 0.11\).
  8. Use the Tables: \(z = -0.84\); then using unstandardising formula, the height is \(x = \mu + (z\times\sigma) = 161.4 + (-0.84\times6.7) = 155.772\), or about \(156\) cm.

Answers for Sect. 6.4

Answers embedded.

Relative frequency. Relative frequency. Subjective. Relative frequency. Classical.

Answers for Sect. 6.5

  1. Depends on data.
  2. Looking close to normal, centred around \(0.5\)-ish.
  3. We'd have an approximate normal distribution with mean \(\mu_{\hat{p}} = 0.5\) and standard deviation \(\text{s.e.}(\hat{p}) = \sqrt{0.5 \times (1 - 0.5)/10} \approx 0.158\).
  4. Still normal; same mean; but standard deviation would be smaller: \(\text{s.e.}(\hat{p}) = \sqrt{0.5 \times (1 - 0.5)/50} \approx 0.071\).

Answers for Sect. 6.6.1

  1. \(z = (20 - 0)/10 = 2\). The area or probability to the right is \(0.0228\), or about \(2.3\)%.
  2. The probability the SOI exceeds \(20\): \(2.3\)%. So odds: \(2.3\div(100 - 2.3)\), or about \(0.024\).
  3. \(z = (-25 - 0)/10 = -2.5\). The area to the left is \(0.0062\), or about \(0.6\)%.
  4. \(z = (-12 - 0)/10 = -1.2\). The area to the right is \(0.8849\), or about \(88.5\)%.
  5. The two \(z\)-scores: \((-10 - 0)/10 = -1\) and \((20 - 0)/10 = 2\). The area between these is \(0.9772 - 0.1587 = 0.8185\), or about \(81.9\)%.
  6. The \(z\) score corresponds to an area of \(0.80\) to the left; from the tables, about \(z = 0.84\). This corresponds to an SOI of \(0 + (0.84\times 10)\), or an SOI of about \(8.4\).
  7. The \(z\)-score is \(0.385\) from Appendix 3 in the textbook, remembering that the area to the left would be \(0.650\) (draw a diagram!). So, the \(z\)-score is \(0.385\), so that the SOI values is \(x = \mu + (z\times\sigma) = 0 + (0.385\times 10) = 3.85\).
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