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 300 apples. p(240 < a < 300) = ?% be sure to use the 68% - 95% - 99.7% rule and do not round

Explanation:

Step1: Recall the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that for a normal distribution:

• Approximately 68% of the data lies within \( \mu\pm\sigma\) (where \(\mu\) is the mean and \(\sigma\) is the standard deviation)
• Approximately 95% of the data lies within \( \mu\pm2\sigma\)
• Approximately 99.7% of the data lies within \( \mu\pm3\sigma\)

We have \(\mu = 300\) and \(\sigma=30\).

Step2: Analyze the intervals

\(240=\mu - 2\sigma\) (since \(300-2\times30 = 240\)) and \(300=\mu\)

The interval from \(\mu - 2\sigma\) to \(\mu\) is half of the interval from \(\mu - 2\sigma\) to \(\mu+ 2\sigma\)

The percentage of data within \(\mu\pm2\sigma\) is 95%.

Step3: Calculate the required probability

The probability \(P(240 < a<300)\) is half of the 95% interval (because the normal distribution is symmetric about the mean \(\mu\)). So \(P(240 < a<300)=\frac{95\%}{2}=47.5\%\)

Answer:

47.5%