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?

Respuesta :

Answer:

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\mu = 8.02, \sigma = 0.4[/tex]

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{8.02 - 8.02}{0.4}[/tex]

[tex]Z = 0[/tex]

[tex]Z = 0[/tex] has a pvalue of 0.5

X = 7.76

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{7.76 - 8.02}{0.4}[/tex]

[tex]Z = -0.65[/tex]

[tex]Z = -0.65[/tex] 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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{7.52 - 8.02}{0.4}[/tex]

[tex]Z = -1.25[/tex]

[tex]Z = -1.25[/tex] has a pvalue of 0.1056

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