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A study was conducted by a team of college students for the college research center. From the study, it was reported that most shoppers have a specific spending limit in place while shopping online. The reports indicate that men spend an average of $230 online before they decide to visit a store. If the spending limit is normally distributed and the standard deviation is $19.
(a) Find the probability that a male spent at least $210 online before deciding to visit a store. Ans: ____________
(b) Find the probability that a male spent between $240 and $300 online before deciding to visit a store. Ans: ____________
(c) Find the probability that a male spent exactly $250 online before deciding to visit a store. Ans: (d) Ninety-one percent of the amounts spent online by a male before deciding to visit a store are less than what value? Ans: ____________

Sagot :

fichoh

Answer:

0.8536

0.29933

Step-by-step explanation:

Given :

Mean amount spent, μ = $230

Standard deviation, σ = $19

1.)

Probability of spending atleast $210

P(x ≥ 210)

The Zscore = (x - μ) / σ = (210 - 230) / 19 = - 1.052

P(Z ≥ -1.052) = 1 - P(Z ≤ - 1.052) = 1 - 0.1464 = 0.8536

2.)

Probability that between $240 and $300 is spent:

P(x < $240) = Zscore = (240 - 230) / 19 = 0.526

P(Z < 0.526) = 0.70056

P(x < 300) = Zscore = (300 - 230) / 19 = 3.684

P(Z < 3.684) = 0.99989

P(Z < 3.684) - P(Z < 0.526)

0.99989-0.70056 = 0.29933

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