Automatic Loader Coal is carried from a mine in West Virginia to a Power plant in New York inhopper cars on a long train. The automatic hopper car loader is set to put 75 tons of coal into each car. Theactual weights of coal loaded into each car are normally distributed, with mean of 75 tons and standard4.deviation of 0.8 ton.(a) State N- ((b) What is the probability that one car chosen at random will have less than 74.5 tons of coal?(c) What is the probability that 20 cars chosen at random will have a mean load weight x of less than 74tons of coal?

Name: Automatic Loader Coal is carried from a mine in West Virginia to a Po
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Description: Automatic Loader Coal is carried from a mine in West Virginia to a Power plant in New York inhopper cars on a long train. The automatic hopper car loader is set to put 75 tons of coal into each car. Theactual weights of coal loaded into each car are

Answered: (a) State N~ ( (b) What is the…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

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Automatic Loader Coal is carried from a mine in West Virginia to a Power plant in New York in
hopper cars on a long train. The automatic hopper car loader is set to put 75 tons of coal into each car. The
actual weights of coal loaded into each car are normally distributed, with mean of 75 tons and standard
4.
deviation of 0.8 ton.
(a) State N- (
(b) What is the probability that one car chosen at random will have less than 74.5 tons of coal?
(c) What is the probability that 20 cars chosen at random will have a mean load weight x of less than 74
tons of coal?

Expert Solution

Step 1

(a)

Let x denote the weights of coal loaded into each car which follows normal distribution with mean 75 and standard deviation 0.8. That is, $N ~ X (75, 0.8)$ .

Thus, the distribution is $N ~ (75, 0.8)$ .

(b)

The probability that one car chosen at random would have less than 74.5 tons of coal is,

$\begin{array}{rcl} P (X < 74.5) & = & P (\frac{X - μ}{σ} < \frac{74.5 - 75}{0.8}) \\ = & P (z < \frac{- 0.5}{0.8}) \\ = & P (z < - 0.625) \end{array}$

The probability of z less than –0.625 can be obtained using the excel formula “=NORM.S.DIST(–0.625,TRUE)”. The probability value is 0.2660.

The required probability value is,

$\begin{array}{rcl} P (X < 74.5) & = & P (z < - 0.625) \\ = & 0.2660 \end{array}$

Thus, the probability that one car chosen at random would have less than 74.5 tons of coal is 0.2660.

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(a) State N~ ( (b) What is the probability that one car chosen at random will have less than 74.5 tons of coal? (c) What is the probability that 20 cars chosen at random will have a mean load weight x of less than 74 tons of coal?