Answer:
0.818
Explanation:
To find the probability that a randomly selected orange from the orchard weighs between 7 oz. and 10 oz., we need to use the properties of the normal distribution. In this case, the distribution is normal with a mean (\(\mu\)) of 8 oz. and a standard deviation (\(\sigma\)) of 1 oz.
The probability that a randomly selected orange falls between 7 oz. and 10 oz. can be calculated by finding the z-scores for 7 oz. and 10 oz., and then using the cumulative normal distribution function (also known as the cumulative distribution function or CDF) to find the area under the normal curve between these z-scores.
Let's calculate the z-scores for 7 oz. and 10 oz. using the formula:
\[z = \frac{x - \mu}{\sigma}\]
For 7 oz.:
\[z_1 = \frac{7 - 8}{1} = -1\]
For 10 oz.:
\[z_2 = \frac{10 - 8}{1} = 2\]
Now, we will use the z-scores to find the cumulative probabilities using a standard normal distribution table or calculator. The area between \(z_1\) and \(z_2\) represents the probability that the orange weighs between 7 oz. and 10 oz.
\[P(7 \leq x \leq 10) = P(-1 \leq z \leq 2)\]
Using a standard normal distribution table or calculator, you can find the values corresponding to \(z_1\) and \(z_2\), and then subtract the cumulative probabilities to find the desired probability.
Keep in mind that values from the standard normal distribution table might need to be converted to cumulative probabilities, depending on the format of the table. The answer will be the area under the normal curve between the two z-scores, representing the probability that a randomly selected orange weighs between 7 oz. and 10 oz.