Question

Suppose that the distribution of the number of defects on any given bolt of cloth is the Poisson distribution with a mean 5, and the number of defects on each bolt is counted for a random sample of 125 bolts. Determine the probability that the average number of defects per bolt in the sample will be less than 5.5.

Short answer

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

Step-by-step solution

Given information

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.

Finding the probability

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