select the correct answer.\nthe average summer temperature in anchorage is 69°f. if the daily temperature is…

select the correct answer.\nthe average summer temperature in anchorage is 69°f. if the daily temperature is normally distributed with a standard deviation of 7°f, what percentage of the time would the temperature be between 55°f and 76°f?\na. 34.0%\nb. 47.5%\nc. 68.0%\nd. 81.5%\ne. 99.7%

select the correct answer.\nthe average summer temperature in anchorage is 69°f. if the daily temperature is normally distributed with a standard deviation of 7°f, what percentage of the time would the temperature be between 55°f and 76°f?\na. 34.0%\nb. 47.5%\nc. 68.0%\nd. 81.5%\ne. 99.7%

Answer

Explanation:

Step1: Calculate z - scores

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value from the data set. For $x = 55$, $\mu=69$, and $\sigma = 7$, $z_1=\frac{55 - 69}{7}=\frac{- 14}{7}=-2$. For $x = 76$, $\mu = 69$, and $\sigma=7$, $z_2=\frac{76 - 69}{7}=\frac{7}{7}=1$.

Step2: Use the standard normal distribution table

The standard - normal distribution table gives the cumulative probability for a given z - score. The cumulative probability for $z=-2$ is $P(Z < - 2)=0.0228$. The cumulative probability for $z = 1$ is $P(Z < 1)=0.8413$.

Step3: Find the probability between the two z - scores

$P(-2<Z<1)=P(Z < 1)-P(Z < - 2)$. $P(-2<Z<1)=0.8413 - 0.0228=0.8185\approx81.5%$.

Answer:

D. 81.5%