Question

According to a government study among adults in the 25- to 34-year age group, the mean amount spent per year on reading and entertainment is $2,060. Assume that the distribution of the amounts spent follows the normal distribution with a standard deviation of $495. (Round your z-score computation to 2 decimal places and final answers to 2 decimal places.)

What percent of the adults spend more than $2,575 per year on reading and entertainment?

What percent spend between $2,575 and $3,300 per year on reading and entertainment?

What percent spend less than $1,225 per year on reading and entertainment?

Answer 1

Answer:

A) P(x> 2575) = 14.92%

B) P(2575 < X<3300) = 14.32%

C)  ( P X < 1225) = 4.55%

Step-by-step explanation:

Given data:

$\mu = 2060$

$\sigma  = 495$

a) P(x> 2575)

 1 - P(X<2575)

$1 - P(\frac{x-\mu}{\sigma} < \frac{2575 - 2060}{495})$

$1 - P(z < \frac{515}{495})$

1 - P( Z< 1.04)

1 - 0.8508

0.1492

14.92%

B) P(2575 < X<3300)

$P(\frac{2575 - 2060}{495} <x < \frac{3300 -2060}{495})$

P (1.04 < z< 2.51)

P(z <2.51) - P(Z<1.04)

0.994 - 0.8508 = 0.1432 = 14.32%

C) ( P X < 1225)

$P( \frac{x - \mu}{\sigma} < \frac{1225 - 2060}{495})$

$P( z < \frac{-835}{495}$

P ( Z< -1.69)

= 0.0455 = 4.55%