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.
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:
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2022/formulas/mathematics/college/bnaa16b36eg8ubb4w75g6u0qutzsb68wqa.png)
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
![\mu = 8.02, \sigma = 0.4](https://img.qammunity.org/2022/formulas/mathematics/college/bl96w6pv7gagbjbf2z3odijm3xhod9z6al.png)
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
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2022/formulas/mathematics/college/bnaa16b36eg8ubb4w75g6u0qutzsb68wqa.png)
![Z = (8.02 - 8.02)/(0.4)](https://img.qammunity.org/2022/formulas/mathematics/college/prqm30y4zjh62j3h7uwyhobfs9ncuh3f67.png)
![Z = 0](https://img.qammunity.org/2022/formulas/mathematics/college/6fbtyd2uqket1rrn9ugije2hmpco8hpyw8.png)
has a pvalue of 0.5
X = 7.76
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2022/formulas/mathematics/college/bnaa16b36eg8ubb4w75g6u0qutzsb68wqa.png)
![Z = (7.76 - 8.02)/(0.4)](https://img.qammunity.org/2022/formulas/mathematics/college/888t650incr5taiv1jkxdfdmt224q61k6m.png)
![Z = -0.65](https://img.qammunity.org/2022/formulas/mathematics/college/n0fbn4po87jc7ufewxehefnilvtrumty5g.png)
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
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2022/formulas/mathematics/college/bnaa16b36eg8ubb4w75g6u0qutzsb68wqa.png)
![Z = (7.52 - 8.02)/(0.4)](https://img.qammunity.org/2022/formulas/mathematics/college/uz04xwhpya8bdhh1kokcuem5q0178b0g71.png)
![Z = -1.25](https://img.qammunity.org/2022/formulas/mathematics/college/1ckoze3lmr16zpsqzh1n6392yd4krp2dm5.png)
has a pvalue of 0.1056
0.1056 = 10.56% probability that a hamburger selected at random with less than 7.52 ounces.