Question

in a certain orchard, the number of apples (a) on a tree is normally distributed with a mean of 300 apples and a standard deviation of 30 apples. find the probability that a given tree has between 240 and 330 apples. p(240 < a < 330) = ?% be sure to use the 68% - 95% - 99.7% rule and do not round.

Explanation:

Step1: Analyze the distance from the mean

The mean is \(300\). For \(240\): \(300 - 240=60\), and since the standard deviation \(\sigma = 30\), \(60\div30 = 2\) standard deviations below the mean. For \(330\): \(330 - 300 = 30\), which is \(1\) standard deviation above the mean.

Step2: Use the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that about \(95\%\) of the data is within \(2\) standard deviations of the mean (\(240\) to \(360\)) and about \(68\%\) of the data is within \(1\) standard deviation of the mean (\(270\) to \(330\)).
The area from \(240\) to \(270\) is \(\frac{95\%}{2}=47.5\%\) (half of the \(95\%\) interval outside the \(68\%\) interval). The area from \(270\) to \(330\) is \(68\%\).

Step3: Calculate the probability

\(P(240\lt a\lt330)=\frac{95\%}{2}+\frac{68\%}{2}\)

$$P(240\lt a\lt330)=47.5\% + 34\%=81.5\%$$

Answer:

\(81.5\)