Mathematics, 05.05.2020 17:07 lamashermosa23

A small publisher wishes to publish self-improvement books. After a survey of the market, the publisher finds that the average cost of the type of book that she wishes to publish is $15.00. If she wants to price her books to sell in the middle 64% range, what should the maximum and minimum prices of the books be? The standard deviation is $0.25 and the variable is normally distributed.

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Answer from: acesmith

Maximum price of book will be $15.23

Minimum price of book will be $14.77.

Step-by-step explanation:

We are given that a small publisher wishes to publish self-improvement books. After a survey of the market, the publisher finds that the average cost of the type of book that she wishes to publish is $15.00. The standard deviation is $0.25 and the variable is normally distributed.

Also, she wants to price her books to sell in the middle 64% range.

The above situation represents that we want 64% confidence interval for the price of book which she wants to publish.

Firstly, the pivotal quantity for 64% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average cost of the type of book = $15

$\sigma$ = population standard deviation = $0.25

n = sample of book = 1

$\mu$ = population mean

Here level of significance = $\frac{1-0.64}{2}$ = 18%

Here for constructing 64% confidence interval we have used One-sample z test statistics because we know about population standard deviation.

So, 64% confidence interval for the population mean, $\mu$ is ;

P(-0.9195 < N(0,1) < 0.9195) = 0.64 {As the critical value of z at 18%

level of significance are -0.9195 & 0.9195}

P(-0.9195 < $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ < 0.9195) = 0.64

P( $-0.9195 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $0.9195 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.64

P( $\bar X-0.9195 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+0.9195 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.64

64% confidence interval for $\mu$ = [ $\bar X-0.9195 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X +0.9195 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $15-0.9195 \times {\frac{0.25}{\sqrt{1} } }$ , $15-0.9195 \times {\frac{0.25}{\sqrt{1} } }$ ]

= [$14.77 , $15.23]

Therefore, the maximum price of book will be $15.23 and the minimum price of book will be $14.77.

Answer from: Quest

yes

step-by-step explanation:

Answer from: Quest

answer: 42

use a calculator to find that 845/73 = 11.575

the whole part of the answer is 11. multiply this by 73 to get 11*73 = 803.

subtract that result from 845 to get 845 - 803 = 42

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