Question

The mean number of words per minute (WPM) typed by a speed typist is 149149 with a standard deviation of 1414 WPM. What is the probability that the sample mean would be greater than 147.8147.8 WPM if 8888 speed typists are randomly selected

Answer 1

The probability is
`P(\= X > x ) = 0.78814`

Explanation:

From the question we are given that

The population mean is
`\mu = 149`

The standard deviation is
`\sigma = 14`

The random number
`x = 147.81`

The sample size is
`n = 88`

The probability that the sample mean would be greater than
`P(\= X > x ) = P( ( \= x - \mu )/(\sigma_(\= x) ) > ( x - \mu )/(\sigma_(\= x) ) )`

Generally the z- score of this normal distribution is mathematically represented as

`Z = ( \= x - \mu )/(\sigma_(\= x) )`

Now

`\sigma_(\= x ) = (\sigma )/(√(n) )`

substituting values

`\sigma_(\= x ) = (14 )/(√(88) )`

`\sigma_(\= x ) = 1.492`

Which implies that

`P(\= X > x ) = P( Z > ( 147.81 - 149 )/( 1.492) )`

`P(\= X > x ) = P( Z > -0.80 )`

Now from the z-table the probability is found to be

`P(\= X > x ) = 0.78814`