Question

Assume the random variable x is normally distributed with mean muμequals=8787 and standard deviation sigmaσequals=55. Find the indicated probability. ​P(x less than<7979​)

Answer 1

Answer: 0.4404

Explanation:

Let the random variable x is normally distributed .

Given : Mean :
`\mu=\ 87`

Standard deviation :
`\sigma= 55`

The formula to calculate the z-score :-

`z=(x-\mu)/(\sigma)`

For x = 79

`z=(79-87)/(55)\approx-0.15`

The p-value =
`P(x<79)=P(z<-0.15)`

`=0.4403823\approx0.4404`

Hence, the required probability :
`P(x<79)=0.4404`