Burger Queen a popular fast food restaurant among college students has undertaken a study to determine the weights of the hamburgers served in its new location in San Marcos, TX. Half pounders should weigh at least 8 ounces. The weight is assumed to be normally distributed. A sample of 500 hamburgers reveals the average precooked weight to be 8.02 ounces, with a standard deviation of .4 ounce. 1. What is the probability that a hamburger selected at random will weigh between 7.76 and 8.02 ounces?
2. What is the probability that a hamburger selected at random with less than 7.52 ounces?

1. 0.2422 = 24.22% probability that a hamburger selected at random will weigh between 7.76 and 8.02 ounces

2. 0.1056 = 10.56% probability that a hamburger selected at random with less than 7.52 ounces.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be 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.

A sample of 500 hamburgers reveals the average precooked weight to be 8.02 ounces, with a standard deviation of .4 ounce.

This means that

1. What is the probability that a hamburger selected at random will weigh between 7.76 and 8.02 ounces?

This is the pvalue of Z when X = 8.02 subtracted by the pvalue of Z when X = 7.76. So

X = 8.02

has a pvalue of 0.5

X = 7.76

has a pvalue of 0.2578

0.5 - 0.2578 = 0.2422

0.2422 = 24.22% probability that a hamburger selected at random will weigh between 7.76 and 8.02 ounces.

2. What is the probability that a hamburger selected at random with less than 7.52 ounces?

This is the pvalue of Z when X = 7.52. So

has a pvalue of 0.1056

0.1056 = 10.56% probability that a hamburger selected at random with less than 7.52 ounces.

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0

translate k to n and reflect across the line containing jk.