assume the random variable x is normally distributed with mean $mu = 86$ and standard deviation $sigma = 4$…

assume the random variable x is normally distributed with mean $mu = 86$ and standard deviation $sigma = 4$. find the indicated probability.\np(74 < x < 81)\np(74 < x < 81)= \n(round to four decimal places as needed.)

assume the random variable x is normally distributed with mean $mu = 86$ and standard deviation $sigma = 4$. find the indicated probability.\np(74 < x < 81)\np(74 < x < 81)= \n(round to four decimal places as needed.)

Answer

Explanation:

Step1: Standardize the bounds

We use the formula $z=\frac{x - \mu}{\sigma}$. For $x = 74$, $z_1=\frac{74 - 86}{4}=\frac{- 12}{4}=-3$. For $x = 81$, $z_2=\frac{81 - 86}{4}=\frac{-5}{4}=-1.25$.

Step2: Rewrite the probability

$P(74 < x<81)=P(-3<z<-1.25)$.

Step3: Use the standard - normal table

$P(-3<z<-1.25)=P(z < - 1.25)-P(z < - 3)$. From the standard - normal table, $P(z < - 1.25)=0.1056$ and $P(z < - 3)=0.0013$.

Step4: Calculate the probability

$P(-3<z<-1.25)=0.1056 - 0.0013=0.1043$.

Answer:

$0.1043$