Question

question
when emily goes bowling, her scores are normally distributed with a mean of 100 and a standard deviation of 10. what percentage of the games that emily bowls does she score between 86 and 125, to the nearest tenth?
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Explanation:

Step1: Identify the distribution parameters

The scores are normally distributed with mean $\mu = 100$ and standard deviation $\sigma = 10$. We need to find $P(86 < X < 125)$.

Step2: Calculate the z - scores

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$.

For $x = 86$:
$z_1=\frac{86 - 100}{10}=\frac{- 14}{10}=-1.4$

For $x = 125$:
$z_2=\frac{125 - 100}{10}=\frac{25}{10}=2.5$

Step3: Find the probabilities from the z - table

We know that $P(86 < X < 125)=P(-1.4 < Z < 2.5)$.

Using the property of the normal distribution $P(-1.4 < Z < 2.5)=P(Z < 2.5)-P(Z < - 1.4)$

From the standard normal table:
$P(Z < 2.5)=0.9938$
$P(Z < - 1.4)=0.0808$

Step4: Calculate the probability

$P(-1.4 < Z < 2.5)=0.9938 - 0.0808 = 0.913$

To convert this to a percentage, we multiply by 100: $0.913\times100 = 91.3\%$

Answer:

91.3%