40 points! Karen created a sample of 20 students from a population of 100 in the following way. First, she gave each student a u
nique number from 1 to 100. She then obtained a list of 20 randomly generated numbers between 1 and 100. She used these numbers to identify which students to select. Here is the list of random numbers:
Which three students is part of her sample?
A. students 1, 10, and 20
B. students 12, 60, and 74
C. students 81, 89, and 90
D. students 3, 10, and 21
Which three students is part of her sample?
A. students 1, 10, and 20
B. students 12, 60, and 74
C. students 81, 89, and 90
D. students 3, 10, and 21
2 answers:
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Answer:
The company should use a mean of 12.37 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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.
The distribution for the amount of beer dispensed by the machine follows a normal distribution with a standard deviation of 0.17 ounce.
This means that
The company can control the mean amount of beer dispensed by the machine. What value of the mean should the company use if it wants to guarantee that 98.5% of the bottles contain at least 12 ounces (the amount on the label)?
This is , considering that when
, Z has a p-value of
, so when
.
Then
The company should use a mean of 12.37 ounces.