Question

If a variable X is normally distributed with a standard deviation of 8, what is the mean value of X, if 0.3413 of the values of X is greater than 58?

Answer 1

Answer:

55

Explanation:

The z score is a score used in statistics to determine how many standard deviations the raw score is above or below the mean.it is given by the formula:

$z=\frac{x-\mu}{\sigma}\\ Where\ \mu \ is\ the\ mean, x\ is\ the\ raw\ score\ and\ \sigma\ is\ the\ standard\ deviation$

Given that:  

σ = 8, P(z > 58) = 0.3413, x > 58/ Hence:

P(z < 58) = 1 - P(z > 58) =  1 - 0.3413 = 0.6587

P(z < 58) = 0.6587

From the normal distribution table, P(z < 58) = 0.6587 corresponds with a z score of 0.41

$z=\frac{x-\mu}{\sigma}\\ \\0.41=\frac{58-\mu}{8}\\\\Cross-multiplying:\\\\58-\mu=3.28\\\\\mu=58-3,28\\\\\mu=54.72$

μ ≅ 55