In a certain orchard, the number of apples or a in 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.

Values at standard deviations negative 3 though 3 are as follows
210 240 270 300 330 360 390

210 to 390 is 99.7%, 240 to 360 is 95%, 270 to 330 is 68%

probability of 240 is less than a which is less than 330 equals [?] percent.
Hint: use the 68 percent 95 percent 99 point 7 percent rule and do not round.

We are given that the number of apples (a) in a tree is normally distributed with a mean (μ) of 300 apples and standard deviation (σ) of 30 apples.

To find the probability that a given tree has between 240 and 330 apples, P(240 < a < 330), we can use the empirical rule (also known as the 68-95-99.7 rule) which states:

Approximately 68% of the data falls within one standard deviation of the mean.
Approximately 95% of the data falls within two standard deviations of the mean.
Approximately 99.7% of the data falls within three standard deviations of the mean.

Since 240 is two standard deviations below the mean and 330 is one standard deviation above the mean, we can consider this range as covering 68% of data and half of the difference between 95% of data and 68% of the data, since the normal distribution curve is symmetrical about the mean.

[tex]\sf P(240 < a < 330)=P(240 < a < 270)+P(270 < a < 330)\\\\\\\sf P(240 < a < 330)=\dfrac{95\%-68\%}{2}+68\%\\\\\\\sf P(240 < a < 330)=\dfrac{27\%}{2}+68\%\\\\\\\sf P(240 < a < 330)=13.5\%+68\%\\\\\\\sf P(240 < a < 330)=81.5\%[/tex]

Therefore, the probability that a given tree has between 240 and 330 apples is approximately:

[tex]\Large\boxed{\boxed{\sf 81.5\%}}[/tex]