Question

1. height is normally distributed with a mean of 67 inches and a standard deviation of 4 inches. what percentage of people have heights between: (6 points)

63 in. and 71 in. a. 34%
b. 95%
67 in. and 75 in. c. 47.5%
d. 49.85%
55 in. and 79 in. e. 99.7%
59 in. and 75 in. f. 68%
g. 83.85%
55 in. and 67 in.
63 in. and 79 in.

Explanation:

Step1: Recall the properties of normal distribution

In a normal - distribution, about 68% of the data lies within 1 standard deviation ($\mu\pm\sigma$) of the mean, about 95% lies within 2 standard deviations ($\mu\pm2\sigma$) of the mean, and about 99.7% lies within 3 standard deviations ($\mu\pm3\sigma$) of the mean. The mean $\mu = 67$ inches and the standard deviation $\sigma=4$ inches.

Step2: Calculate the number of standard - deviations for each range

For 63 in. and 71 in.:

The value 63 is $\frac{67 - 63}{4}=1$ standard deviation below the mean and 71 is $\frac{71 - 67}{4}=1$ standard deviation above the mean. So the percentage of data between 63 and 71 is 68% (f).

For 67 in. and 75 in.:

67 is the mean. 75 is $\frac{75 - 67}{4}=2$ standard deviations above the mean. The percentage of data between the mean and 2 standard deviations above the mean is $\frac{95\%}{2}=47.5\%$ (c).

For 55 in. and 79 in.:

55 is $\frac{67 - 55}{4}=3$ standard deviations below the mean and 79 is $\frac{79 - 67}{4}=3$ standard deviations above the mean. So the percentage of data between 55 and 79 is 99.7% (e).

For 59 in. and 75 in.:

59 is $\frac{67 - 59}{4}=2$ standard deviations below the mean and 75 is 2 standard deviations above the mean. The percentage of data between 2 standard deviations below and 2 standard deviations above the mean is 95% (b).

For 55 in. and 67 in.:

55 is 3 standard deviations below the mean and 67 is the mean. The percentage of data between 3 standard deviations below the mean and the mean is $\frac{99.7\%}{2}=49.85\%$ (d).

For 63 in. and 79 in.:

63 is 1 standard deviation below the mean and 79 is 3 standard deviations above the mean. The percentage of data between 1 standard deviation below the mean and 3 standard deviations above the mean is $\frac{68\%+99.7\%}{2}=83.85\%$ (g).

Answer:

1. 63 in. and 71 in. - f. 68%
2. 67 in. and 75 in. - c. 47.5%
3. 55 in. and 79 in. - e. 99.7%
4. 59 in. and 75 in. - b. 95%
5. 55 in. and 67 in. - d. 49.85%
6. 63 in. and 79 in. - g. 83.85%