Question

A hot dog vendor sells an average of 50 hot dogs during a Little League baseball game. If the sales are Normally distributed with a standard deviation of 7 hot dogs, what is the probability the vendor will sell between 45 and 65 hot dogs? Show your work.

Answer 1

Answer:

74.64%

Explanation:

Average sales (μ) = 50 hot dogs

Standard deviation (σ) = 7 hot dogs

In a normal distribution, the z-score for any given number of hot dogs sold, X, is determined by:

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

For X = 45 hot dogs:

$z=\frac{45-50}{7}\\ z= -0.7143$

For X = 65 hot dogs:

$z=\frac{65-50}{7}\\ z= 2.1429$

A z-score of -0.7143 falls in the 23.75th percentile of a normal distribution while a z-score of 2.1429 falls in the 98.39th percentile.

Therefore, the probability that he vendor will sell between 45 and 65 hot dogs is:

$P(45 \leq X \leq 65) = 98.39-23.75\\P(45 \leq X \leq 65) = 74.64\%$