Question

One year consumers spent an average of

​$21

on a meal at a resturant. Assume that the amount spent on a resturant meal is normally distributed and that the standard deviation is

​$5.

Complete parts​ (a) through​ (c) below.

a.

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

​$22?

​P(Xgreater than>​$22​)=

​(Round to four decimal places as​ needed.)

b.

What is the probability that a randomly selected person spent between

​$9

and

​$18?

​P($9less than<Xless than<​$18​)equals=nothing

​(Round to four decimal places as​ needed.)

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

Xequals=​

and

Xequals=​.

Answer 1

I have answered the question below

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Answer:

a).

Since μ = 21 and σ = 5 we have:

P ( X > 22 ) = P ( X−μ > 22−21 ) = P ( (X−μ)/σ > (22−21)/5)

Since Z = (x−μ)/σ and (22−21)/5 = 0.2 we have:

P ( X > 22 ) = P ( Z > 0.2 )

Using the standard normal table to conclude that:

P (Z > 0.2) = 0.4207

b).

Since μ = 21 and σ = 5 we have:

P ( 9 < X < 18 ) = P ( 9−21 < X−μ < 18−21 ) = P ( (9−21)/5 < (X−μ)/σ < (18−21)/5)

P ( 9 < X < 18 ) = P ( −2.4 < Z < −0.6 )

Using the standard normal table to conclude that:

P ( −2.4 < Z < −0.6 ) = 0.2661

c).Thus, lower limit of X =

= 21-1.96*5 = 11.2

Upper limit of X = = 30.8

The middle 95% of the amounts of cash spent will fall between X = $11.2 and X = $30.8