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.

Respuesta :

Answer:

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. = [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]  ~ N(0,1)

where, [tex]\bar X[/tex] = sample average cost of the type of book = $15

            [tex]\sigma[/tex] = population standard deviation = $0.25

            n = sample of book = 1

            [tex]\mu[/tex] = population mean

Here level of significance =  [tex]\frac{1-0.64}{2}[/tex] = 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, [tex]\mu[/tex] 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 < [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 0.9195) = 0.64

P( [tex]-0.9195 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]{\bar X -\mu}[/tex] < [tex]0.9195 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.64

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

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

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

                                       = [$14.77 , $15.23]

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

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