Question

for questions 10 - 12, refer to the situation below
assume that the heights of college women are normally distributed, with mean 60 inches and standard deviation 1.5 inches.
10 what percentage of women is taller than 64.5 inches?
a. 16%
b. 2.5%
c. 2.35%
d. 0.15%

1. what percentage of women is shorter than 57 inches?

a. 16%
b. 2.5%
c. 2.35%
d. 0.15%

1. what percentage of women is between 55.5 inches and 63 inches?

a. 99.7%
b. 97.5%
c. 97.35%
d. 95%
use the following information for questions 13 & 14:
iq test scores are normally distributed with a mean of 100 and a standard deviation of 15.

1. an individuals iq score is found to be 110. find the z - score corresponding to this value

a. 0.67
b. - 1.33
c. - 0.67
d. 1.33

Explanation:

Step1: Calculate z - score for question 10

The formula for z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 64.5\), \(\mu=60\), \(\sigma = 1.5\).
\(z=\frac{64.5 - 60}{1.5}=\frac{4.5}{1.5}=3\)
Using the empirical rule (68 - 95 - 99.7 rule), for a normal distribution, about \(99.7\%\) of the data lies within \(z=- 3\) and \(z = 3\). The percentage of data above \(z = 3\) is \(\frac{100 - 99.7}{2}=0.15\%\)

Step2: Calculate z - score for question 11

For \(x = 57\), \(\mu=60\), \(\sigma = 1.5\)
\(z=\frac{57 - 60}{1.5}=\frac{-3}{1.5}=-2\)
Using the empirical rule, the percentage of data below \(z=-2\) is \(\frac{100 - 95}{2}=2.5\%\)

Step3: Calculate z - scores for question 12

For \(x_1 = 55.5\), \(z_1=\frac{55.5 - 60}{1.5}=\frac{-4.5}{1.5}=-3\)
For \(x_2 = 63\), \(z_2=\frac{63 - 60}{1.5}=\frac{3}{1.5}=2\)
The percentage of data between \(z=-3\) and \(z = 2\) is \(49.85+47.5 = 97.35\%\) (since between \(z = 0\) and \(z=3\) is \(49.85\%\) and between \(z = 0\) and \(z = 2\) is \(47.5\%\))

Step4: Calculate z - score for question 13

For \(x = 110\), \(\mu = 100\), \(\sigma=15\)
\(z=\frac{110 - 100}{15}=\frac{10}{15}\approx0.67\)

Answer:

1. D. \(0.15\%\)
2. B. \(2.5\%\)
3. C. \(97.35\%\)
4. A. \(0.67\)