Question

NOTE!! Please provide the solution in MS Excell.  

One year consumers spent an average of $21 on a meal at a restaurant. Assume that the amount spent on a restaurant meal is normally distributed and that the standard deviation is $33

Complete parts (a) through (c) below.

a.

What is the probability that a randomly selected person spent more than $22?

b.

What is the probability that a randomly selected person spent between $14 and $20

c.

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

Answer 1

Solution:-

Mean = 21

S.D = 33

a) The probability that a randomly selected person spent more than $22 is 0.488.

x = 22

By applying normal distruibution:-

z = 0.03

P(z > 0.03) = 0.488

b) The probability that a randomly selected person spent between $14 and $20 is  0.0718.

x₁ = 14

x₂ = 20

By applying normal distruibution:-

z₁ = - 0.212

z₂ = - 0.03

P( -0.212 < z < - 0.0303) = P(z > - 0.212) - P(z > - 0.0303)

P( -0.212 < z < - 0.0303) = 0.5839 - 0.5121

P( -0.212 < z < - 0.0303) = 0.0718

c) The two values will the middle 95% of the amounts of cash spent fall is - 43.68 and 85.68.

p-value for the middle 95% = 0.025 and 0.975

z-score for the p-value = + 1.96

By applying normal distruibution:-

x₁ = - 43.68

x₂ = 85.68