Question

Isabella averages 152 points per bowling game with a standard deviation of 14.5 points. Suppose Isabella's points per bowling game are normally distributed. Let X= the number of points per bowling game. Then X∼N(152,14.5)______.
If necessary, round to three decimal places.
Suppose Isabella scores 187 points in the game on Sunday. The z-score when x=187 is ___ The mean is _________
This z-score tells you that x = 187 is _________ standard deviations.

Answer 1

Answer:

The z-score when x=187 is 2.41. The mean is 187. This z-score tells you that x = 187 is 2.41 standard deviations above the mean.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

$\mu = 152, \sigma = 14.5$

The z-score when x=187 is ...

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{187 - 152}{14.5}$

$Z = 2.41$

The z-score when x=187 is 2.41. The mean is 187. This z-score tells you that x = 187 is 2.41 standard deviations above the mean.