Question

1. an individuals iq score is found to be 120. find the z - score corresponding to this value.

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

1. assume that salaries of elementary school teachers are normally distributed with a mean of $32,000 and a standard deviation of $3,000. find the probability a teacher makes less than $28,000

a. 9.18% b. 98.27% c. 21.13% d. 99.81%

1. if heights of 3rd graders follow a normal distribution with a mean of 52 inches and a standard deviation of 2.5, what is the z - score of a 3rd grader who is 47 inches tall?

a. - 5 b. - 2 c. 2 d. 5

1. a national achievement test is administered annually to 3rd graders. the test has a mean score of 100 and a standard deviation of 15. if janes z - score is 1.20, what was her score on the test?

a. 82 b. 88 c. 100 d. 118

Explanation:

Step1: Recall the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step2: Solve problem 14

Assume for IQ scores, \(\mu = 100\) and \(\sigma=15\). Given \(x = 120\), then \(z=\frac{120 - 100}{15}=\frac{20}{15}\approx1.33\)

Step3: Solve problem 15

Given \(\mu = 32000\), \(\sigma=3000\), \(x = 28000\). First, calculate \(z=\frac{28000 - 32000}{3000}=\frac{- 4000}{3000}\approx - 1.33\). Then, using the standard normal table, \(P(Z\lt - 1.33)=0.0918 = 9.18\%\)

Step4: Solve problem 16

Given \(\mu = 52\), \(\sigma=2.5\), \(x = 47\). Then \(z=\frac{47 - 52}{2.5}=\frac{-5}{2.5}=-2\)

Step5: Solve problem 17

Given \(z = 1.20\), \(\mu = 100\), \(\sigma=15\). Using the formula \(z=\frac{x-\mu}{\sigma}\), we can rewrite it as \(x=\mu+z\sigma\). Substitute the values: \(x=100+1.20\times15=100 + 18=118\)

Answer:

1. C. \(1.33\)
2. A. \(9.18\%\)
3. B. \(-2\)
4. D. \(118\)