Question

coal is carted 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 89 tons of coal into each car. the actual weights of coal loaded into each car are normally distributed, with mean μ = 89 tons and standard deviation σ = 0.7 ton. (a) what is the probability that one car chosen at random will have less than 88.5 tons of coal? (round your answer to four decimal places.) (b) what is the probability that 26 cars chosen at random will have a mean load weight x of less than 88.5 tons of coal? (round your answer to four decimal places.) (c) suppose the weight of coal in one car was less than 88.5 tons. would that fact make you suspect that the loader had slipped out of adjustment? suppose the weight of coal in 26 cars selected at random had an average x of less than 88.5 tons. would that fact make you suspect that the loader had slipped out of adjustment? why? o yes, the probability that this deviation is random is very large. o yes, the probability that this deviation is random is very small. o no, the probability that this deviation is random is very small. o no, the probability that this deviation is random is very large.

Explanation:

Step1: Calculate the z - score for part (a)

The formula for the z - score is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). Given \(\mu = 89\), \(\sigma=0.7\), \(n = 1\), and \(\bar{x}=88.5\).

$$ z=\frac{88.5 - 89}{\frac{0.7}{\sqrt{1}}}=\frac{- 0.5}{0.7}\approx - 0.7143 $$

Using the standard normal distribution table (or a calculator with a normal - distribution function), \(P(X<88.5)=\Phi(-0.7143)\)

$$ P(X < 88.5)=0.2375 $$

Step2: Calculate the z - score for part (b)

Now \(n = 26\). The formula for the z - score is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). Substitute \(\mu = 89\), \(\sigma = 0.7\), \(n = 26\), and \(\bar{x}=88.5\)

$$ z=\frac{88.5-89}{\frac{0.7}{\sqrt{26}}}=\frac{-0.5}{\frac{0.7}{5.099}}\approx\frac{-0.5}{0.1373}\approx - 3.6416 $$

Using the standard normal distribution table (or a calculator with a normal - distribution function), \(P(\bar{X}<88.5)=\Phi(-3.6416)\)

$$ P(\bar{X}<88.5)=0.0001 $$

Step3: Answer part (c)

For a single car (\(n = 1\)), a load of \(88.5\) tons is not extremely unusual (\(z\approx - 0.71\)), but for a sample of \(n = 26\) cars, a sample mean of \(88.5\) tons is extremely unusual (\(z\approx-3.64\)). Since the probability that the mean load weight of 26 cars is less than \(88.5\) tons is very small (\(P = 0.0001\)), we would suspect that the loader had slipped out of adjustment.

Answer:

(a) \(0.2375\)
(b) \(0.0001\)
(c) Yes, because the probability that the mean of 26 cars is less than \(88.5\) tons is very small.