Question

A normally distributed population has a mean of 76 and a standard deviation of 19. Determine the probability that a random sample of size 22 has an average between 72 and 76.

Round to four decimal places.

Answer 1

Solution :

Given that,

mean = $\mu$ = 76

standard deviation = $\sigma$ =19

n=22

$\mu$_{$\bar x$} = 76

$\sigma$_{$\bar x$} =$\sigma$  / $\sqrt$ n = 19/ $\sqrt$ 22=4.0508

= P(72< $\bar x$    < 76) = P[(72 - 76) / 4.0508< ( $\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ $\bar x$ < (76- 76) /4.0508 )]

= P( -0.99< Z <0 )

= P(Z <0 ) - P(Z <-0.99 )

Using z table,  

= 0.5 - 0.1611

= 0.3389