68.

- Use the
*z*-score formula.*z*= –0.5141. The height of 77 inches is 0.5141 standard deviations below the mean. An NBA player whose height is 77 inches is shorter than average. - Use the
*z*-score formula.*z*= 1.5424. The height 85 inches is 1.5424 standard deviations above the mean. An NBA player whose height is 85 inches is taller than average. - Height = 79 + 3.5(3.89) = 92.615 inches, which is taller than 7 feet, 8 inches. There are very few NBA players this tall so the answer is no, not likely.

72.

Let *X* = an SAT math score and *Y* = an ACT math score.

*X*= 720 $\frac{\text{720\u2013520}}{\text{15}}$ = 1.74 The exam score of 720 is 1.74 standard deviations above the mean of 520.*z*= 1.5

The math SAT score is 520 + 1.5(115) ≈ 692.5. The exam score of 692.5 is 1.5 standard deviations above the mean of 520.- $\frac{X\text{\u2013}\mu}{\text{\sigma}}$ = $\frac{\text{700\u2013514}}{\text{117}}$ ≈ 1.59, the
*z*-score for the SAT. $\frac{Y\text{\u2013}\mu}{\sigma}$ = $\frac{30\text{\u2013}21}{5.3}$ ≈ 1.70, the*z*-scores for the ACT. With respect to the test they took, the person who took the ACT did better (has the higher*z*-score).

79.

*X*~*N*(36, 10)- The probability that a person consumes more than 40% of their calories as fat is 0.3446.
- Approximately 25% of people consume less than 29.26% of their calories as fat.

81.

*X*= number of hours that a Chinese four-year-old in a rural area is unsupervised during the day.*X*~*N*(3, 1.5)- The probability that the child spends less than one hour a day unsupervised is 0.0918.
- The probability that a child spends over ten hours a day unsupervised is less than 0.0001.
- 2.21 hours

83.

*X*= the distribution of the number of days a particular type of criminal trial will take*X*~*N*(21, 7)- The probability that a randomly selected trial will last more than 24 days is 0.3336.
- 22.77

85.

- mean = 5.51,
*s*= 2.15 - Check student's solution.
- Check student's solution.
- Check student's solution.
*X*~*N*(5.51, 2.15)- 0.6029
- The cumulative frequency for less than 6.1 minutes is 0.64.
- The answers to part f and part g are not exactly the same, because the normal distribution is only an approximation to the real one.
- The answers to part f and part g are close, because a normal distribution is an excellent approximation when the sample size is greater than 30.
- The approximation would have been less accurate, because the smaller sample size means that the data does not fit normal curve as well.

88.

*n*= 100;*p*= 0.1;*q*= 0.9*μ*=*np*= (100)(0.10) = 10*σ*= $\sqrt{npq}$ = $\sqrt{\text{(100)(0}\text{.1)(0}\text{.9)}}$ = 3

- $z=\pm 1:{x}_{1}=\mu +\mathrm{z\sigma}=10+\mathrm{1(3)}=13$ and $x2=\mu \u2013\mathrm{z\sigma}=10\u2013\mathrm{1(3)}=\mathrm{7.68\%}$ of the defective cars will fall between seven and 13.
- $z=\pm 2:{x}_{1}=\mu +\mathrm{z\sigma}=10+\mathrm{2(3)}=16$ and $x2=\mu \u2013\mathrm{z\sigma}=10\u2013\mathrm{2(3)}=\mathrm{4.\; 95\; \%}$ of the defective cars will fall between four and 16
- $z=\pm 3:{x}_{1}=\mu +\mathrm{z\sigma}=10+\mathrm{3(3)}=19$ and $x2=\mu \u2013\mathrm{z\sigma}=10\u2013\mathrm{3(3)}=\mathrm{1.\; 99.7\%}$ of the defective cars will fall between one and 19.

90.

*n*= 190;*p*= \frac{1}{5} = 0.2;*q*= 0.8*μ*=*np*= (190)(0.2) = 38*σ*= $\sqrt{npq}$ = $\sqrt{\text{(190)(0}\text{.2)(0}\text{.8)}}$ = 5.5136

- For this problem:
*P*(34 <*x*< 54) = 0.7641 - For this problem:
*P*(54 <*x*< 64) = 0.0018 - For this problem:
*P*(*x*> 64) = 0.0000012 (approximately 0)