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

2. $z = (20 - 0)/10 = 2$. The area or probability to the right is $0.0228$, or about $2.3$%.
3. The probability the SOI exceeds $20$: $2.3$%. So odds: $2.3\div(100 - 2.3)$, or about $0.024$.
4. $z = (-25 - 0)/10 = -2.5$. The area to the left is $0.0062$, or about $0.6$%.
5. $z = (-12 - 0)/10 = -1.2$. The area to the right is $0.8849$, or about $88.5$%.
6. 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 $81.9$%.
7. 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 $8.4$.
8. 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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