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2 answers:
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Correct Answer:
Option A. 0.01
Solution:
This is a problem of statistics and uses the concept of normal distributions. We need to convert the score of 90 into z-score and then find the desired probability from standard normal distribution table.
Converting 90 to z-score:
Now we are to find the probability of z score being more than 2.33. From the z-table the probability comes out to be 0.01.
Therefore, we can conclude that the probability of class average is greater than 90 is 0.01.
Option A. 0.01
Solution:
This is a problem of statistics and uses the concept of normal distributions. We need to convert the score of 90 into z-score and then find the desired probability from standard normal distribution table.
Converting 90 to z-score:
Now we are to find the probability of z score being more than 2.33. From the z-table the probability comes out to be 0.01.
Therefore, we can conclude that the probability of class average is greater than 90 is 0.01.
Answer: 0.22
Step-by-step explanation:
We know that the best point estimate for the difference between two population mean is the difference between their sample means.
Given : For the 39 randomly selected upperclassmen, the sample mean was 0.12 and sample standard deviation was 0.42. For the 35 randomly selected underclassmen, the sample mean was 0.34 and the sample standard deviation was 0.87.
Let A denotes the population of upperclassmen and B denotes the population of underclassmen .
Then, the point estimate of the difference in the population mean volunteered between underclassmen and upperclassmen will be :-
Hence, the point estimate of the difference in the population mean volunteered between underclassmen and upperclassmen =0.22