Renee scores an average of 153 points in a game of bowling, and her points in a game of bowling are normally distributed. Suppose Renee scores 175 points in a game, and this value has a z-score of 2. What is the standard deviation? Do not include the units in your answer. For example, if you found that the standard deviation was 15 points, you would enter 15.

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".

Let X the random variable that represent points in a game of bowling of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(153,unknown)[/tex]

Where [tex]\mu=153[/tex] and [tex]\sigma=?[/tex]

We know that the value for Renee is X=175 and the z score obteined was Z=2.

Solution to the problem

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

We are interested on the value of [tex]\sigma[/tex] and we can solv for it:

[tex] z \sigma = X-\mu[/tex]

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

And replacing the values we have:

[tex]\sigma=\frac{175-153}{2}=11[/tex]

pamateriales pamateriales · Answer 2 · 2019-12-13T19:11:56+01:00

pamateriales pamateriales

13-12-2019

Answer:

The standard deviation is 11

Step-by-step explanation:

Hi, you need to solve for "S" (standard deviation) the following equation.

[tex]Z=\frac{X-Mean}{S}[/tex]

Where: Z = z-score value, X=Renne´s score, Mean= average value (153)

So, it should look like this

[tex]S=\frac{X-Mean}{Z}[/tex]

Therefore

[tex]S=\frac{175-153}{2}=11[/tex]

So, the standard deviation is 11

Best of luck.

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