Question

1.

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 76 tons of coal into each car. The actual weights of coal loaded into each car are normally distributed, with mean μ = 76 tons and standard deviation σ = 0.5 ton.

(a) What is the probability that one car chosen at random will have less than 75.5 tons of coal? (Round your answer to four decimal places.)


(b) What is the probability that 20 cars chosen at random will have a mean load weight x of less than 75.5 tons of coal? (Round your answer to four decimal places.)


(c) Suppose the weight of coal in one car was less than 75.5 tons. Would that fact make you suspect that the loader had slipped out of adjustment?

YesNo     

Suppose the weight of coal in 20 cars selected at random had an average x of less than 75.5 tons. Would that fact make you suspect that the loader had slipped out of adjustment? Why?

Yes, the probability that this deviation is random is very small.

Yes, the probability that this deviation is random is very large.     

No, the probability that this deviation is random is very small.

No, the probability that this deviation is random is very large.

2. Anystate Auto Insurance Company took a random sample of 362 insurance claims paid out during a 1-year period. The average claim paid was $1535. Assume σ = $230.

Find a 0.90 confidence interval for the mean claim payment. (Round your answers to two decimal places.)

lower limit	$
upper limit	$

Find a 0.99 confidence interval for the mean claim payment. (Round your answers to two decimal places.)

lower limit	$
upper limit	$

Answer 1

1. $\mu = 76, \ \sigma = 0.5$

a) Probability that one car chosen at random will have less than 75.5 tons of coal, P(X<75.5) =

$=P(X<75.5) = P\left (\frac{X-\mu}{\sigma}< \frac{75.5-\mu}{\sigma} \right )$

$= P\left (Z< \frac{75.5-76}{0.5} \right )$

$= P\left (Z<-1 \right )$

From standard normal table:

$=\textbf{0.1587}$

b) Probability that 20 cars chosen at random will have a mean load weight x of less than 75.5 tons of coal, P($\bar X$ < 75.5) =

$=P(\bar X<75.5) = P\left (\frac{\bar X-\mu}{\sigma/\sqrt{n}}< \frac{75.5-\mu}{\sigma/\sqrt{n}} \right )$

$= P\left (Z< \frac{75.5-76}{0.5/\sqrt{20}} \right )$

$= P\left (Z<-4.47 \right )$

From standard normal table:

$=\textbf{0.0000}$

c) i)Suppose the weight of coal in one car was less than 75.5 tons.

No, it would not make us suspect that the loader had slipped out of adjustment.

ii) Suppose the weight of coal in 20 cars selected at random had an average x of less than 75.5 tons.

Yes, the probability that this deviation is random is very small.

2. $\bar X = 1535,\ \sigma=230,\ n=362$

0.90 confidence interval for the mean claim payment

Critical z-value for $\alpha$= 0.1 is z_c = 1.645

$\textup{Lower limit =} \bar X - \frac{z_c*\sigma}{\sqrt{n}} = 1535-\frac{1.645*230}{\sqrt{362}} = \textbf{1515.11}$

$\textup{Upper limit =} \bar X &plus; \frac{z_c*\sigma}{\sqrt{n}} = 1535&plus;\frac{1.645*230}{\sqrt{362}} = \textbf{1554.89}$

0.99 confidence interval for the mean claim payment

Critical z-value for $\alpha$= 0.01 is z_c = 2.576

$\textup{Lower limit =} \bar X - \frac{z_c*\sigma}{\sqrt{n}} = 1535-\frac{2.576*230}{\sqrt{362}} = \textbf{1503.86}$

$\textup{Upper limit =} \bar X &plus; \frac{z_c*\sigma}{\sqrt{n}} = 1535&plus;\frac{2.576*230}{\sqrt{362}} = \textbf{1566.14}$