College

According to the American Dental Association, 8% of adults have never had a cavity. A dental graduate student contacts an SRS of 1000 adults and calculates the proportion [tex]p[/tex] in this sample who have never had a cavity.

1. Identify the mean of the sampling distribution of [tex]p[/tex].
2. Calculate and interpret the standard deviation of the sampling distribution of [tex]p[/tex].
3. Check that the 10% condition is met.
4. Is the sampling distribution of [tex]p[/tex] approximately Normal? Check that the Large Counts condition is met.
5. Find the probability that the random sample of 1000 adults will give a result within 2 percentage points of the true value.
6. If the sample size were 9000 rather than 1000, how would this change the sampling distribution of [tex]p[/tex]?

Answer :

The sample distribution's mean is 8%, or 0.08, and the standard deviation is given number by - = 8.58 n*p = 1000*0.

In its most basic form, a ratio is a comparison between two comparable quantities. There are two types of proportions One is the direct proportion, whereby increasing one number by a constant k also increases the other quantity by the same constant k, and vice versa. If one quantity is increased by a constant k, the other will decrease by the same constant k in the case of inverse proportion, and vice versa.

8% of adults are known to have never had a cavity. An srs of 1000 adults is contacted by a dentistry graduate student.

8% of 1000 is,\s= (8/100)×1000.\s= 80.

Now, The percentage of people who have never had a cavity is (1000 - 80)/100, which translates to the probabilty of 920/1000, 92/100, or 0.92.

To learn more about probabilty from the link: https://brainly.com/question/30034780

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