63.

- Use the
*z*-score formula.*z*= â€“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.

67.

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

*X*= 720 $\frac{\text{720\xe2\u20ac\u201c520}}{\text{15}}$ = 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) â‰ˆ 692.5. The exam score of 692.5 is 1.5 standard deviations above the mean of 520.- $\frac{X\text{\xe2\u20ac\u201c}\mathrm{\xce\xbc}}{\text{\xcf\u0192}}$ = $\frac{\text{700\xe2\u20ac\u201c514}}{\text{117}}$ â‰ˆ 1.59, the
*z*-score for the SAT. $\frac{Y\text{\xe2\u20ac\u201c}\mathrm{\xce\xbc}}{\mathrm{\xcf\u0192}}$ = $\frac{30\text{\xe2\u20ac\u201c}21}{5.3}$ â‰ˆ 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).

75.

*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.

77.

*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

79.

*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

81.

- 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.

83.

- mean = 60,136

*s*= 10,468 - Answers will vary.
- Answers will vary.
- Answers will vary.
*X*~*N*(60136, 10468)- 0.7440
- The cumulative relative frequency is 43/60 = 0.717.
- The answers for part f and part g are not the same, because the normal distribution is only an approximation.

85.

*n*= 100;*p*= 0.1;*q*= 0.9*Î¼*=*np*= (100)(0.10) = 10*Ïƒ*= $\sqrt{npq}$ = $\sqrt{\text{(100)(0}\text{.1)(0}\text{.9)}}$ = 3

*z*= Â±1:*x*=_{1}*Âµ*+*zÏƒ*= 10 + 1(3) = 13 and*x*2 =*Âµ*â€“*zÏƒ*= 10 â€“ 1(3) = 7. 68% of the defective cars will fall between seven and 13.*z*= Â±2:*x*=_{1}*Âµ*+*zÏƒ*= 10 + 2(3) = 16 and*x*2 =*Âµ*â€“*zÏƒ*= 10 â€“ 2(3) = 4. 95 % of the defective cars will fall between four and 16*z*= Â±3:*x*=_{1}*Âµ*+*zÏƒ*= 10 + 3(3) = 19 and*x*2 =*Âµ*â€“*zÏƒ*= 10 â€“ 3(3) = 1. 99.7% of the defective cars will fall between one and 19.

87.

*n*= 190;*p*= \frac{1}{5} = 0.2;*q*= 0.8*Î¼*=*np*= (190)(0.2) = 38*Ïƒ*= $\sqrt{npq}$ = $\sqrt{\text{(190)(0}\text{.2)(0}\text{.8)}}$ = 5.5136

- For this problem:
*P*(34 <*x*< 54) = normalcdf(34,54,48,5.5136) = 0.7641 - For this problem:
*P*(54 <*x*< 64) = normalcdf(54,64,48,5.5136) = 0.0018 - For this problem:
*P*(*x*> 64) = normalcdf(64,10^{99},48,5.5136) = 0.0000012 (approximately 0)