question\nthe weight of oranges growing in an orchard is normally distributed with a mean weight of 8 oz…

question\nthe weight of oranges growing in an orchard is normally distributed with a mean weight of 8 oz. and a standard deviation of 1 oz. what is the probability that a randomly selected orange from the orchard weighs between 7 oz. and 10 oz., to the nearest thousandth?\nstatistics calculator\nanswer attempt 1 out of 2

question\nthe weight of oranges growing in an orchard is normally distributed with a mean weight of 8 oz. and a standard deviation of 1 oz. what is the probability that a randomly selected orange from the orchard weighs between 7 oz. and 10 oz., to the nearest thousandth?\nstatistics calculator\nanswer attempt 1 out of 2

Answer

Explanation:

Step1: Calculate z - scores

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value from the data set. For $x = 7$, $z_1=\frac{7 - 8}{1}=- 1$ For $x = 10$, $z_2=\frac{10 - 8}{1}=2$

Step2: Use the standard normal distribution table

We want to find $P(-1<Z<2)$. We know that $P(-1<Z<2)=P(Z < 2)-P(Z<-1)$. From the standard - normal distribution table, $P(Z < 2)=0.9772$ and $P(Z<-1)=0.1587$. So $P(-1<Z<2)=0.9772 - 0.1587=0.8185\approx0.819$

Answer:

$0.819$