A study found that consumers spend an average of $21 per week in cash without be

Question

A study found that consumers spend an average of $21 per week in cash without being aware of where it goes. Assume that the amount of cash spent without being aware of where it goes is normally distributed and that the standard deviation is $33. Complete parts (a) through (c).

a. What is the probability that a randomly selected person will spend more than $22? P(Xgreater than>$22)equals= .(Round to four decimal places as needed.)

b. What is the probability that a randomly selected person will spend between $13 and $19?

c. Between what two values will the middle 95% of the amounts of cash spent fall?

The middle 95% of the amounts of cash spent will fall between X =$___ and X =$___ (round to nearest cent as needed)

Explanation / Answer

  z = (X-Mean)/SD
a) z = (22-21)/33 = + 0.303
The area under the standard normal curve right to the z value indicates the required probability.
P(X>22) = P(z > 0.303) = 0.6190

b) z1 = (13-21)/33 = -0.2424
z2 = (19-21)/33 = - 0.0606
The area under the standard normal curve between these two z values indicates the required probability.
P(10 < X < 20) = P(-0.2424 < z < 0- 0.0606)
The area under the sta)
= 0.4042 (area corresponding to z1) - 0.4758 (area corrresponding to z2)
= -0.0753

c) Middle 95% is represented by 0.9500 area about mean. It implies that 0.9500/2 = 0.4750 area lies on left side and 0.4750 area lies on right side of mean.
The z value corresponding to 0.4750 area is 1.96
Therefore, Required X values are Mean +/- (z*SD)
21 +/- (1.96*33)
21 +/- 64.68
Middle 95% of the amounts of cash spent will fall between 43.6 and 85.6

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