91Ó°ÊÓ

The number of defects on any given bolt of cloth is a Poisson distribution with a mean of 5, and the number of defects on each bolt is counted for a random sample of 125 bolts.

The number of defects on any bolt has a Poisson distribution with mean\(\lambda = 5\)and variance,\(\lambda = 5\)

Therefore, the distribution of the average number\({\bar X_n}\)on the 125 bolts will be approximately the normal distribution, with the mean being,

\(\mu = 5\)

And the variance is,

\(\begin{array}{c}{\sigma ^2} = \frac{5}{{125}}\\ = \frac{1}{{25}}\end{array}\)

The standard deviation is,

\(\begin{array}{c}\sigma = \sqrt {\frac{1}{{25}}} \\ = \frac{1}{5}\end{array}\)

Let,

\(\begin{array}{c}Z = \frac{{\left( {{{\bar X}_n} - \mu } \right)}}{\sigma }\\ = \frac{{{{\bar X}_n} - 5}}{{\frac{1}{5}}}\\ = 5\left( {{{\bar X}_n} - 5} \right)\end{array}\)

Then the distribution of Z will be an approximately standard normal distribution.

\({\bar X_n} = 5.5\)

\(\begin{array}{c}Z = \frac{{\left( {{{\bar X}_n} - \mu } \right)}}{\sigma }\\ = \frac{{{{\bar X}_n} - 5}}{{\frac{1}{5}}}\\ = 5\left( {5.5 - 5} \right)\\ = 2.5\end{array}\)

The probability is,

\(\begin{array}{c}P\left( {{{\bar X}_n} < 5.5} \right) = P\left( {Z < 2.5} \right)\\ = 0.9937903\\ \approx 0.9938\end{array}\)

Therefore, the probability that the average number of defects per bolt in the sample is less than 5.5 is 0.9938