Question

The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer randomly selected 36 swans and loaded them into his truck. What is the probability that this flock of swans weighs > 1000 pounds?

Answer 1

Let
`X` denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by
`X_1,\ldots,X_(36)`, each independently and identically distributed with distribution
`X_i\sim\mathcal N(26,7.2)`.

You want to find


`\mathbb P(X_1+\cdots+X_(36)>1000)=\mathbb P\left(\displaystyle\sum_(i=1)^(36)X_i>1000\right)`

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to


`\mathbb P\left(36\displaystyle\sum_(i=1)^(36)(X_i)/(36)>1000\right)=\mathbb P\left(\overline X>(1000)/(36)\right)`

Recall that if
`X\sim\mathcal N(\mu,\sigma)`, then the sampling distribution
`\overline X=\displaystyle\sum_(i=1)^n\frac{X_i}n\sim\mathcal N\left(\mu,\frac\sigma{\sqrt n}\right)` with
`n` being the size of the sample.

Transforming to the standard normal distribution, you have


`Z=(\overline X-\mu_(\overline X))/(\sigma_(\overline X))=\sqrt n(\overline X-\mu)/(\sigma)`

so that in this case,


`Z=6(\overline X-26)/(7.2)`

and the probability is equivalent to


`\mathbb P\left(\overline X>(1000)/(36)\right)=\mathbb P\left(6(\overline X-26)/(7.2)>6((1000)/(36)-26)/(7.2)\right)`

`=\mathbb P(Z>1.481)\approx0.0693`