Question

The weight of oranges growing in an orchard is normally distributed with a mean

weight of 4 oz. and a standard deviation of 0.5 oz. From a batch of 2000 oranges, how
many would be expected to weight less than 5 oz., to the nearest whole number?

Answer 1

The weight of oranges growing in an orchard is normally distributed with a mean weight of 6.5 oz. and a standard deviation of 1 oz. Using the empirical rule, determine what interval would represent weights of the middle 99.7% of all oranges from this orchard

Explanation:

bran leist ple

Answer 2

1954

Explanation:

If a continuous random variable X is normally distributed with mean μ and variance σ²:

`\boxed{X \sim \textsf{N}(\mu,\sigma^2)}`

Given:

• mean μ = 4 oz
• standard deviation σ = 0.5 oz

First find the probability that the weight of an orange is less than 5 oz.

`\text{If \;$X \sim \textsf{N}(4,0.5^2)$,\;\;find\;\;P$(X < 5)$}.`

Method 1

Using a calculator:

`\implies \text{P}(X < 5)=0.977249868`

Method 2

Converting to the z-distribution.

`\boxed{\text{If\;\;$X \sim$N$(\mu,\sigma^2)$\;\;then\;\;$(X-\mu)/(\sigma)=Z$, \quad where $Z \sim$N$(0,1)$}}`

`x=5 \implies Z=(5-4)/(0.5)=2`

Using the z-tables for the probability:

`\implies \text{P}(Z < 2)=0.9772`

To find the expected number of oranges that will weigh less than 5 oz from a batch of 2000, multiply the total number of oranges by the probability calculated:

`\begin{aligned}2000 * \text{P}(X < 5)&=2000 * 0.977249868\\&=1954.499736\\&=1954\end{aligned}`

Therefore, 1954 oranges would be expected to weigh less than 5 oz.