Question

A publisher has discovered that the numbers of words contained in a new manuscript are normally distributed with a mean equal to 20,000 words in excess of that specified in the author's contract and a standard deviation of 10,000 words. If the publisher wants to be almost certain (say, witha probability of 0.95) that the manuscript will have less than 100,000 words, what number of words should the publisher specify the contract?​

Answer 1

Answer:

2000 .will less than 100,000

Answer 2

Given:

The normally distributed with a mean the is equal to $20,000$ a standard deviation of $10,000$ words in the manuscript will have less than$100,000$.

To find:

The number of words the publisher specify the contract.

Step-by-step explanation:

The x be the number of words in the manuscript and c be the words in the contract.

X~N$($μ$C+20000$,$x=10000$)

Determining the z-score corresponding to a probability of $95$% or$0.95$ in a standard normal table as $1.645$.

From the above expression,

$z_{x}=\frac{100000-c-20000}{10000}$

$1.645=\frac{100000-(c+20000}{10000}$

$1.645=\frac{80000-c}{10000}$

$16450=80000-c$

$c=80000-16450$

$c=63550$

Answer:

Therefore, The count specified in the contract must be $63550$.