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.)

Question

asked Aug 8, 2020 217k views

1 Answer

Zeantsoi · Answer 1 · 2020-08-12T20:38:03+0000

Answer:

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)

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%.