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?

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:

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%

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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 and standard deviation , the zscore of a measure X is given by:

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:

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

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

has a pvalue of 0.2119.

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

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)

how am i supposed to answer

step-by-step explanation: