Question

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

a) - 5
b) - 2
c) 2
d) 5

1. the mean temperature in glens falls for the month of february is 23 degrees with a standard deviation of 4.2 degrees. what is the z - score for a temperature of 17 degrees?

a) 1
b) 1.43
c) - 1.43
d) 11.5

1. a data set has a mean of 300 and a standard deviation of 40. what value would have a z - score of z = - 2?

a) 298
b) 296
c) 220

Explanation:

Step1: Recall the z - score formula

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

Step2: Substitute values for problem 3

For problem 3, \(x = 47\), \(\mu=52\), \(\sigma = 2.5\). Then \(z=\frac{47 - 52}{2.5}=\frac{-5}{2.5}=- 2\).

Step3: Substitute values for problem 4

For problem 4, \(x = 17\), \(\mu = 23\), \(\sigma=4.2\). Then \(z=\frac{17 - 23}{4.2}=\frac{-6}{4.2}\approx - 1.43\).

Step4: Rearrange the z - score formula for problem 5

From \(z=\frac{x-\mu}{\sigma}\), we can solve for \(x\) as \(x=\mu+z\sigma\). Given \(z=-2\), \(\mu = 300\), \(\sigma = 40\), then \(x=300+(-2)\times40=300 - 80=220\).

Answer:

1. b) - 2
2. c) - 1.43
3. c) 220