Question

A population of values has a normal distribution with μ = 101.4 and σ = 82.4 . You intend to draw a random sample of size n = 129 .

Find the probability that a single randomly selected value is greater than 96.3. P(X > 96.3) =

Find the probability that a sample of size n = 129 is randomly selected with a mean greater than 96.3. P( ¯ x > 96.3)=

Enter your answers as numbers accurate to 4 decimal places.

Answer 1

Solution :

Given that ,

mean = $\mu$ = 101.4

standard deviation = $\sigma$ = 82.4

P(x >96.3 ) = 1 - P(x <96.3 )

= 1 - P[(x - $\mu$ ) / $\sigma$ < (96.3 - 101.4) /82.4 ]

= 1 - P(z < -0.06)

Using z table,

= 1 -0.4761

=0.5239

(B)

n = 129

$\mu$_{$\bar x$} = 101.4

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 82.4 / $\sqrt$ 129 = 7.2549

P($\bar x$ > 96.3) = 1 - P($\bar x$ < 96.3)

= 1 - P[($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (96.3 - 101.4) / 7.2549]

= 1 - P(z <-0.70 )

Using z table,    

= 1 - 0.242

= 0.7580