Question

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the weight of oranges growing in an orchard is normally distributed with a mean weight of 8 oz. and a standard deviation of 1.5 oz. what is the probability that a randomly selected orange from the orchard weighs more than 10 oz., to the nearest thousandth?
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answer attempt 2 out of 2
0.092
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Explanation:

Step1: Identify the distribution and parameters

The weight of oranges is normally distributed with mean $\mu = 8$ oz and standard deviation $\sigma = 1.5$ oz. We need to find $P(X > 10)$, where $X$ is the weight of an orange.

Step2: Calculate the z-score

The z-score formula is $z=\frac{X - \mu}{\sigma}$. Substituting $X = 10$, $\mu = 8$, and $\sigma = 1.5$:
$$z=\frac{10 - 8}{1.5}=\frac{2}{1.5}\approx1.33$$

Step3: Find the probability using z-table

We need $P(X > 10)=P(Z > 1.33)$. Since the total area under the standard normal curve is 1, $P(Z > 1.33)=1 - P(Z \leq 1.33)$. From the z-table, $P(Z \leq 1.33)\approx0.9082$. So, $P(Z > 1.33)=1 - 0.9082 = 0.0918\approx0.092$ (to the nearest thousandth).

Answer:

0.092