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coldgirl [10]
3 years ago
13

Average sales for an online textbook distributor were $67.63 per customer per purchase. Assume the sales are normally distribute

d. If the standard deviation of the amount spent on textbooks is $8.96, find these probabilities for a randomly selected customer of the online textbook distributor. Round your answer to the nearest tenth of a percent. (a)He or she spent more than $85.55 per purchase. (b)He or she spent less than $76.59 per purchase
Mathematics
1 answer:
Fynjy0 [20]3 years ago
6 0

Answer:

(a)P(x>85.55)=0.02275

(b)P(x

Step-by-step explanation:

We are given that

Average sales for an online textbook distributor  per customer per purchase

,\mu=$67.63

Standard deviation of the amount spent on textbooks,\sigma=$8.96

(a) We have to find probability for a randomly selected customer spent more than $85.55 per purchase.

P(x>85.55)=P(\frac{x-\mu}{\sigma}>\frac{85.55-67.63}{8.96})

=P(Z>2)

=1-P(Z\leq 2)

=1-0.97725

P(x>85.55)=0.02275

(b)We have to find probability for a randomly selected customer spent less than $76.59 per purchase

P(x

P(x

P(x

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Step-by-step explanation:

Let's work backward from that 81%:  x/25 = 0.81 yields x = 20.25.  Nominally, 20.25 / 25 = 0.81, but x must be an integer.  Let's round 20.25 off to 20.

Thus, if Kalsom got 81%, it was a result of his having done 20 questions correctly.

81% corresponds to 20 questions correct;

82% to 20.5 questions correct, or, rounding up, to 21 questions correct;

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84% to 21 questions correct; this is the only result that makes sense (whole number of questions answered correctly)

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Assuming that the number of questions correct MUST be integer, then the possible number correct are 21 and 22, corresponding to 84% and 88% respectively.

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