Question

The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about ten. Suppose that 16 individuals are randomly chosen.

(a) For the group of 16, find the probability that the average percent of fat calories consumed is more than thirty-five (Round to 3 decimal places)

(b) Find the first quartile for the percent of fat calories. (Round to 4 decimal places)

(c) Find the first quartile for the average percent of fat calories. (Round to 3 decimal places)

Answer 1

Solution :

mean = $\mu$ = 36

standard deviation = $\sigma$ = 10

n = 16

$\mu$_{$\bar x$} =  $\mu$ = 36

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 10 / $\sqrt$ 16 = 2.5

a) P($\bar x$ > 35) = 1 - P($\bar x$ < 35)

= 1 - P[($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (35 - 36) / 2.5 ]

= 1 - P(z < -0.40)

Using z table,    

= 1 - 0.345

= 0.655

b) The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.6745

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = -0.6745 * 10 + 36

x = 29.255

First quartile =Q1 = 29.255

c) The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.6745

Using z-score formula  

$\bar x$ = z * $\sigma$ _{$\bar x$}+ $\mu$ _{$\bar x$}

$\bar x$ = -0.6745 * 2.5 + 36

$\bar x$ = 34.314

First quartile =Q1 = 34.314