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 1.5 oz. what percentage of the oranges from the orchard weigh between 5 oz. and 7 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 weight of oranges is normally distributed with mean $\mu = 7.5$ oz and standard deviation $\sigma = 1.5$ oz. We need to find $P(5 < X < 7)$, 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 $x = 5$:
$z_1=\frac{5 - 7.5}{1.5}=\frac{- 2.5}{1.5}\approx - 1.67$

For $x = 7$:
$z_2=\frac{7 - 7.5}{1.5}=\frac{-0.5}{1.5}\approx - 0.33$

Step3: Find the probabilities using the standard normal distribution

We know that $P(5 < X < 7)=P(-1.67 < Z < - 0.33)$, where $Z$ is a standard normal variable.

We also know that $P(-1.67 < Z < - 0.33)=P(Z < - 0.33)-P(Z < - 1.67)$

Using the standard normal table or a calculator:

• $P(Z < - 0.33)\approx0.3707$
• $P(Z < - 1.67)\approx0.0475$

Step4: Calculate the probability

$P(5 < X < 7)=0.3707 - 0.0475 = 0.3232$

To convert this to a percentage, we multiply by 100: $0.3232\times100 = 32.32\%\approx32.3\%$

Answer:

$32.3\%$