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 1 oz. from a batch of 1200 oranges, how many would be expected to weight more than 5 oz., to the nearest whole number?
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answer attempt 1 out of 2
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Explanation:

Step1: Calculate z-score

$z = \frac{x - \mu}{\sigma} = \frac{5 - 7.5}{1} = -2.5$

Step2: Find upper tail probability

$P(X > 5) = P(Z > -2.5) = 1 - P(Z \leq -2.5)$
Using standard normal table, $P(Z \leq -2.5) = 0.0062$, so $P(X > 5) = 1 - 0.0062 = 0.9938$

Step3: Calculate expected number

$\text{Number} = 1200 \times 0.9938 = 1192.56$

Step4: Round to nearest whole number

$\text{Rounded value} = 1193$

Answer:

1193