in a certain orchard, the number of apples (a) on a tree is normally distributed with a mean of 300 apples…

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. 99.7% 95% 68% 210 240 270 300 330 360 390 p(240 < a < 300) = ?% be sure to use the 68% - 95% - 99.7% rule and do not round.

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. 99.7% 95% 68% 210 240 270 300 330 360 390 p(240 < a < 300) = ?% be sure to use the 68% - 95% - 99.7% rule and do not round.

Answer

Explanation:

Step1: Calculate the z - scores

The mean $\mu = 300$ and the standard deviation $\sigma=30$. For $x = 240$, the z - score $z=\frac{x - \mu}{\sigma}=\frac{240 - 300}{30}=\frac{- 60}{30}=-2$. The mean has a z - score of $z = 0$ (since $x=\mu$ gives $z=\frac{\mu-\mu}{\sigma}=0$).

Step2: Apply the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that about 95% of the data lies within 2 standard deviations of the mean, i.e., $P(-2<z<2) = 95%$. The normal distribution is symmetric about the mean. So the probability between $z=-2$ and $z = 0$ is half of the probability between $z=-2$ and $z = 2$.

Answer:

47.5%