Question

Steve Goodman, production foreman for the Florida Gold Fruit Company, estimates that the average sale of oranges is 4,700 and the standard deviation is 500 oranges. Sales follow a normal distribution. What is the probability that sales will be less than 4,300 oranges?

Answer 1

Answer:

21.19% probability that sales will be less than 4,300 oranges.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 4700, \sigma = 500$

What is the probability that sales will be less than 4,300 oranges?

This is the pvalue of Z when X = 4300. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4300 - 4700}{500}$

$Z = -0.8$

$Z = -0.8$ has a pvalue of 0.2119.

So there is a 21.19% probability that sales will be less than 4,300 oranges.

Answer 2

The probability that sales will be less than 4,300 oranges is 21.19%

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

$z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score, \mu=mean, \sigma=standard\ deviation\\\\Given\ that:\\\mu=4700,\sigma=500\\\\For\ x=4300:\\\\z=\frac{4300-4700}{500} =-0.8$

From the normal distribution table:

P(x < 4300) = P(z < -0.8) = 0.2119 = 21.19%

The probability that sales will be less than 4,300 oranges is 21.19%

Find out more on z score at: https://brainly.com/question/25638875