when jayden goes bowling, his scores are normally distributed with a mean of 115 and a standard deviation of…

when jayden goes bowling, his scores are normally distributed with a mean of 115 and a standard deviation of 11. what is the probability that the next game jayden bowls, his score will be higher than 122, to the nearest thousandth?

when jayden goes bowling, his scores are normally distributed with a mean of 115 and a standard deviation of 11. what is the probability that the next game jayden bowls, his score will be higher than 122, to the nearest thousandth?

Answer

Explanation:

Step1: Calculate the z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 122$, $\mu=115$, and $\sigma = 11$. $z=\frac{122 - 115}{11}=\frac{7}{11}\approx0.636$

Step2: Find the probability

We want $P(X>122)$, which is equivalent to $P(Z > 0.636)$ in the standard normal distribution. Since the total area under the standard - normal curve is 1, and $P(Z>z)=1 - P(Z\leq z)$. Looking up $P(Z\leq0.636)$ in the standard normal table (or using a calculator with a normal - distribution function), we find that $P(Z\leq0.636)\approx0.737$. Then $P(Z > 0.636)=1 - 0.737 = 0.263$.

Answer:

$0.263$