Section 1.3 Solutions

 

Note:  These solutions do not include normal curve pictures, but pictures are a very important part of the solution process.

 

1.83

The median line should divide the density curve into two equal pieces (in terms of area under the curve). Hence, the median lines are B, A, and B, respectively. The mean and median will be the same for symmetric density curves, and for skewed curves the mean will be less than the median (for skewed left curves) or greater than the median (for skewed right curves). Hence, the mean lines are C, A, and A, respectively.

 

 

1.87

The lengths of horse pregnancies are approximately N(336 days, 3 days).

 

  1. Almost all (99.7%) of pregnancies fall within 3 standard deviations of the mean. That is, between 336 – 3(3) = 336 – 9 =327 days and 336 + 9 = 345 days.

 

  1. The value of 339 is 1 standard deviation above the mean. Because 68% of the pregnancies are between 333 days and 339 days, only 32% of the pregnancies are less than 333 days or greater than 339 days. By symmetry of the normal curve, 32/2 = 16% of pregnancies are greater than 339 days.

 

 

1.88

The proportions of people in these samples who stay home for fear of crime are N(0.4, 0.015).

 

  1. The mean proportion is 0.4, so by symmetry of the normal curve, 50% of samples have sample proportions above 0.4. The value 0.43 is two standard deviations above the mean. Because 95% of samples produce proportions between 0.37 and 0.43, only 5% of samples produce proportions less than 0.37 or greater than 0.43. By symmetry of the normal curve, 5/2 = 2.5% of samples produce proportions above 0.43.

 

  1. As mentioned in the solution to part a, 95% of samples produce proportions between 0.37 and 0.43.

 

 

Information for the next 4 solutions:

ACT scores are approximately N(20.8, 4.8)

SAT scores are approximately N(1026, 209)

 

1.98

We need to compare the z-scores:

 

 

Tonya scored 1.40 standard deviations above the mean, and Jermaine scored 1.29 standard deviations above the mean. Hence, Tonya has the higher score.

 

 

1.104

 

From Table A, the area above z = 2.75 is (1 - 0.9970) = 0.003. Hence, 0.3% of the SAT scores are reported as 1600.

 

 

1.106

First we need to determine the z-score that corresponds to the top 10%. Because Table A gives us area to the left, we need to do reverse look-up for the area value 0.9. The corresponding z-value is 1.28. Hence, the corresponding SAT score is 1.28 standard deviations above the mean:

x = 1.28(209) + 1026 = 1293.52

 

 

1.108

First we need to find the z-score corresponding to the first quartile, so we must do reverse look-up for the area value 0.25. The corresponding z-value is -0.67. Hence, the first quartile for the ACT scores is 0.67 standard deviations below the mean:

Q1 = -0.67(4.8) + 20.8 = 17.584

 

Then we need to find the z-score corresponding to the third quartile, so we must do reverse look-up for the area value 0.75. The corresponding z-value is 0.67. Hence, the third quartile for the ACT scores is 0.67 standard deviations above the mean:

Q1 = 0.67(4.8) + 20.8 = 24.016

 

 

1.114

The distribution of yearly returns is approximately N(8.3%, 20.3%).

 

  1. About 95% of the yearly returns will be within 2 standard deviations of the mean. That is, within 8.3 – 2(20.3) = -32.3% and 8.3 + 2(20.3) = 48.9%

 

  1. The z-score is , so the value of 0% is 0.41 standard deviations below the mean. Hence (using Table A), we know the market is down 34.09% (or 0.3409) of the years.
  2. The z-score is , so the value of 25% is 0.82 standard deviations above the mean. Hence (using Table A), we know the market gains more than 25% in 20.61% (i.e., 1 – 0.7939 = 0.2061) of the years.

 

 

1.115

The lengths of human pregnancies are approximately N(266 days, 16 days).

 

  1. The z-score is , so the value of 240 days is 1.63 standard deviations below the mean. Hence (using Table A), only 5.16% of pregnancies last less than 240 days.

 

  1. For the value 270, the z-score is . Thus, 59.87% – 5.16% = 54.71% of pregnancies last between 240 and 270 days.

 

  1. The longest 20% of pregnancies are at least 0.84 standard deviations above the mean (from reverse look-up of the area value 0.80 in Table A), or at least 266 + 0.84(16) = 279.44 days.