Question

a. Student GPAs: Bob’s z-score z = + 1.71 μ = 2.98 σ = 0.36 b. Weekly work hours: Sarah’s z-score z = + 1.18 μ = 21.6 σ = 7.1 c. Bowling scores: Dave’s z-score z = - 1.35 μ = 150 σ = 40 Find the original data value corresponding to each standardized z-score. (Round your answers to 2 decimal places.)

a. Bob’s GPA
b. Sarah’s weekly work hours
c. Dave’s bowling score

Answer 1

Answer:

a) $x = \mu +z*\sigma$

And replacing we got:

$x= 2.98 + 1.71*0.36 = 3.5956$

b) $x = \mu +z*\sigma$

And replacing we got:

$x= 21.6 + 1.18*7.1 = 29.978$

c) $x = \mu -z*\sigma$

And replacing we got:

$x= 150 - 1.35*40= 96$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable of interest for this case. We define the z score with the following formula:

$z=\frac{x-\mu}{\sigma}$

And for this case we know that $z = 1.71, \mu = 2.98,\sigma = 0.36$

If we solve for x from the z score formula we got:

$x = \mu +z*\sigma$

And replacing we got:

$x= 2.98 + 1.71*0.36 = 3.5956$

Part b

Let X the random variable of interest for this case. We define the z score with the following formula:

$z=\frac{x-\mu}{\sigma}$

And for this case we know that $z = 1.18, \mu = 21.6,\sigma = 7.1$

If we solve for x from the z score formula we got:

$x = \mu +z*\sigma$

And replacing we got:

$x= 21.6 + 1.18*7.1 = 29.978$

Part c

Let X the random variable of interest for this case. We define the z score with the following formula:

$z=\frac{x-\mu}{\sigma}$

And for this case we know that $z = -1.35, \mu = 150,\sigma = 40$

If we solve for x from the z score formula we got:

$x = \mu -z*\sigma$

And replacing we got:

$x= 150 - 1.35*40= 96$