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3 years ago

7

As the number of pages in a photo book increases, the prices of the book also increases. There is an additional shipping charge

a 15%. The price of a book can be modeled by the equation below, where P= the price of the book, 20 is the printing charge, 0.5 is the charge per page, and x= the number of pages:
P=(20+0.5x) + 0.15(20+0.5x)
Jennifer wants to purchase a book but only has $62.10 to spend. What is the maximum number of pages she can have in her book

1 answer:

Brums [2.3K]3 years ago

6 0

Equate P as 62.10 since we are looking for the MAX value for x (then a legitimate one if necessary)

(20 + 0.5x) + (3 + 0.075x) = 62.1
= 0.575x + 23 = 62.1
x = (62.1-23)/0.575
x = 680

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From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

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Convert this to a normal distribution,

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(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

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Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

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(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

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(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

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$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

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For this case,

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Let $p$ be the probability that one bearing meets the specification. So

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On solving

$p=0.9053$

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