A manager of a department store is interested in determining the percentage of p

Question

A manager of a department store is interested in determining the percentage of people attracted to his store by magazine advertising. Doing an extensive survey, he finds that the average number of people in his store on any given day who have seen the advertising is 200, while the standard deviation of this group is 40 Assuming that the number of people in the store who have seen the advertising has a normal distribution. a) Determine the probability that between 155 and 220 people will see the advertisement. b) Determine the probability that less than 230 people will see the advertisement.

Explanation / Answer

Mean = 200

Standard deviation = 40

a) Probability that between 155 and 220 people will see the advertisement = P(155<X<220)

= P(X<220) - P(X<155)

= P(Z < (220 - 200)/40) - P(Z < (155-200)/40)

= P(Z < 0.5) - P( Z < -1.125)

= 0.6915 - 0.1303

= 0.5612

b) Probability that less than 230 people will see the advertisement = P(X<230)

= P(Z < (230-200)/40)

= P(Z < 0.75)

= 0.7734

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