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 330 and 390 apples. p(330 < a < 390) = ?% 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 \( \mu\pm3\sigma\)

Here, \(\mu = 300\) and \(\sigma=30\)

Step2: Analyze the intervals

For \(330\): \(330=\mu+\sigma\) (since \(300 + 30=330\))
For \(390\): \(390=\mu + 3\sigma\) (since \(300+3\times30=300 + 90 = 390\))

The interval \(330\lt a\lt390\) can be written as \((\mu+\sigma)\lt a\lt(\mu + 3\sigma)\)

We know that the interval \(\mu - 3\sigma\lt a\lt\mu+3\sigma\) contains 99.7% of the data and the interval \(\mu-\sigma\lt a\lt\mu+\sigma\) contains 68% of the data.

The interval \(\mu+\sigma\lt a\lt\mu + 3\sigma\) is half of the interval \((\mu - 3\sigma\lt a\lt\mu+3\sigma)\) minus half of the interval \((\mu-\sigma\lt a\lt\mu+\sigma)\)

The proportion of data in \(\mu - 3\sigma\lt a\lt\mu+3\sigma\) is \(P(\mu - 3\sigma\lt a\lt\mu+3\sigma)=99.7\%\)

The proportion of data in \(\mu-\sigma\lt a\lt\mu+\sigma\) is \(P(\mu-\sigma\lt a\lt\mu+\sigma)=68\%\)

The proportion of data in \(\mu+\sigma\lt a\lt\mu + 3\sigma=\frac{99.7\% - 68\%}{2}\)

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Answer:

\(15.85\)