Answer:
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
Step-by-step explanation:
bran leist ple
Answer:
1954
Step-by-step explanation:
If a continuous random variable X is normally distributed with mean μ and variance σ²:
[tex]\boxed{X \sim \textsf{N}(\mu,\sigma^2)}[/tex]
Given:
First find the probability that the weight of an orange is less than 5 oz.
[tex]\text{If \;$X \sim \textsf{N}(4,0.5^2)$,\;\;find\;\;P$(X < 5)$}.[/tex]
Method 1
Using a calculator:
[tex]\implies \text{P}(X < 5)=0.977249868[/tex]
Method 2
Converting to the z-distribution.
[tex]\boxed{\text{If\;\;$X \sim$N$(\mu,\sigma^2)$\;\;then\;\;$\dfrac{X-\mu}{\sigma}=Z$, \quad where $Z \sim$N$(0,1)$}}[/tex]
[tex]x=5 \implies Z=\dfrac{5-4}{0.5}=2[/tex]
Using the z-tables for the probability:
[tex]\implies \text{P}(Z < 2)=0.9772[/tex]
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:
[tex]\begin{aligned}2000 \times \text{P}(X < 5)&=2000 \times 0.977249868\\&=1954.499736\\&=1954\end{aligned}[/tex]
Therefore, 1954 oranges would be expected to weigh less than 5 oz.
