Question

One year consumers spent an average of ​$22 on a meal at a restaurant. Assume that the amount spent on a restaurant meal is normally distributed and that the standard deviation is ​$4 Complete parts​ (a) through​ (c) below.

a. What is the probability that a randomly selected person spent more than $24

b. What is the probability that a randomly selected person spent between $12 and $19

c. Middle 95% of the amounts of cash spent will fall between X = $___ and X = $___

Answer 1

Solution :

Given that ,

a) P(x > 24) = 1 - p( x< 24)

=1- p P[(x - ) / < (24 - 22) / 4]

=1- P(z < 0.50 )

= 1 - 0.6915

= 0.3085

b) P(12 < x < 19) = P[(12 - 22)/ 4) < (x - ) /  < (19 - 22) / 4) ]

= P(-2.50 < z < -0.75)

= P(z < -0.75) - P(z < -2.50)

Using z table,

= 0.2266 - 0.0062

= 0.2204

c) Using standard normal table,

P( -z < Z < z) = 95%

= P(Z < z) - P(Z <-z ) = 0.95

= 2P(Z < z) - 1 = 0.95

= 2P(Z < z) = 1 + 0.95

= P(Z < z) = 1.95 / 2

= P(Z < z) = 0.975

= P(Z < 1.96) = 0.975

= z  ± 1.96

Using z-score formula,

x = z * +

x = -1.96 * 4 + 22

x = 14.16

Using z-score formula,

x = z * +

x = 1.96 * 4 + 22

x = 29.84

The middle 95% are from $ 14.16 to $ 29.84