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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.

Respuesta :

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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:

[tex]\mu = 152, \sigma = 14.5[/tex]

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{187 - 152}{14.5}[/tex]

[tex]Z = 2.41[/tex]

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.

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