Question

A population has a normal distribution with a mean of 51.5 and a standard deviation of 9.6. Assuming , the probability, rounded to four decimal places, that the sample mean of a sample of size 23 elements selected from this populations will be more than 51.15 is:

Answer 1

Solution :

Given that,

mean = = 51.5

standard deviation = = 9.6

n=23

= =51.5

= / n = 9.6 / 23 = 2.0017

P( >51.15 ) = 1 - P( <51.15 )

= 1 - P[( - ) / < (51.15-51.5) / 2.0017 ]

= 1 - P(z <-0.17 )

Using z table

= 1 - 0.4325

= 0.5675

probability= 0.5675