Question

question
the weight of oranges growing in an orchard is normally distributed with a mean weight of 7.5 oz. and a standard deviation of 0.5 oz. what percentage of the oranges from the orchard weigh between 8 oz. and 9 oz., to the nearest tenth?
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answer attempt 1 out of 2
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Explanation:

Step1: Identify the distribution and parameters

The weights of oranges are normally distributed with mean $\mu = 7.5$ oz and standard deviation $\sigma = 0.5$ oz. We need to find $P(8 < X < 9)$, where $X$ is the weight of an orange.

Step2: Calculate the z - scores

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$.

For $x = 8$:
$z_1=\frac{8 - 7.5}{0.5}=\frac{0.5}{0.5}=1$

For $x = 9$:
$z_2=\frac{9 - 7.5}{0.5}=\frac{1.5}{0.5}=3$

Step3: Find the probabilities using the standard normal table

We know that $P(8 < X < 9)=P(1 < Z < 3)$, where $Z$ is a standard normal variable.

From the standard normal table, $P(Z < 3)\approx0.9987$ and $P(Z < 1)\approx0.8413$.

Then $P(1 < Z < 3)=P(Z < 3)-P(Z < 1)=0.9987 - 0.8413 = 0.1574$

Step4: Convert to percentage and round

To convert to a percentage, we multiply by 100: $0.1574\times100 = 15.74\%$

Rounding to the nearest tenth, we get $15.7\%$

Answer:

$15.7\%$