Question

A study found that consumers spend an average of $22 per week in cash without being aware of where it goes. Assume that the amount of cash spent without beir aware of where it goes is normally distributed and that the standard deviation is $6. Complete parts (a) through (c). a What is the probability that a randomly selected person will spend more than $23 PX>$23)(Round to four decimal places as needed b. What is the probability that a randomly selected person will spend between $6 and $17? Ps6<X<$17)-(Round to four decimal places as needed) C. Between what two values will the midle 95% of the amounts of cash spent far? The middle 95% of the amounts of cash spent will fall between X-S□and X S□ Round to the nearest cent as needed)

Answer 1

(a)

$\mu$ = 22

$\sigma$ = 6

To find P(X>23):
Z = (23 - 22)/6 = 0.1667

Table of Area Under Standard Normal Curve gives area = 0.0675

So,

P(X>23) = 0.5 - 0.0675 = 0.4325

(b)

$\mu$ = 22

$\sigma$ = 6

To find P(6<X>17):

Case 1: For X from 6 to mid value:
Z = (6 - 22)/6 = - 2.6667

Table of Area Under Standard Normal Curve gives area = 0.4962

Case 2: For X from 17 to mid value:

Z = (17 - 22)/ 6 = - 0.8333

Table gives area = 0.2967

So,

P(6 < X <17) = 0.4962 - 0.2967 = 0.1995

(c)

Middle 95% corresponds to area = 0.95/2 = 0.475 on either side of mid value.

Table gives Z = $\pm$ 1.96

Low side:

Z = - 1.96 = (X - 22)/6

So,

X = 22 - (1.96 X 6) = 22 - 11.76 = 10.24

High side:

Z = 1.96 = (X - 22)/6

So,

X = 22 + (1.96 X 6) = 22 + 11.76 = 33.76

So,

The middle 95% of the amount of cash spent will be between X = $10.24 and X = $33.76.