Question

Suppose a variable has a normal distribution with mean 67 and standard deviation 4. What percentage of the distribution is less than 75? (Use z-score.)

Answer 1

The percentage of the distribution is less than 75 is 97.72%.

Explanation:

Given,

Mean of the distribution,

`\mu=67`

Standard deviation,

`\sigma = 4`

Thus, the z-score of the score 75,

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

`=(75-67)/(4)`

`=(8)/(4)`

`=2`

With the help of z-score table,

`P(x<75)=0.9772=97.72\%`

Hence, the percentage of the distribution is less than 75 is 97.72%.