A recent survey found that 72% of all adults over 50 wear glasses for driving. In a random sample of 100 adults over 50, what is the mean and standard deviation of the numbe who wear glasses?

A group of students were asked if they carry a credit card. The responses are listed in the table.

Class Credit Card No credit card total
Freshman 46 14 60
Sophmore 32 8 40
Total 78 22 100

If a student is selected at random, find the probability that he or she owns a credit card given that the student is a sophomore. Round your answer to three decimal places.

0.150

0.060

0.850

0.353

A sports analyst records the winners of NASCAR Winston Cup races for a recent season. The random variable x represents the races won by a driver in one season. Use the frequency distribution to construct a probability distribution.


wine 1 2 3 4 5 6 7
Drivers 12 2 0 2 0 0 1

Gdyd

mean: 72; standard deviation: 8.49

A recent survey found that 70% of all adults over 50 wear glasses for driving. In a random sample of 10 adults over 50, what is the probability that exactly eight wear glasses?

A recent survey found that 63% of all adults over 50 wear glasses for driving. In a random sample of 100 adults over 50, what is the mean and standard deviation of those that wear glasses?


mean: 37; standard deviation: 7.94

mean: 37; standard deviation: 4.83

mean: 63; standard deviation: 7.94

mean: 63; standard deviation: 4.83

A tourist in Ireland wants to visit seven different cities. If the route is randomly selected, what is the probability that the tourist will visit the cities in alphabetical order? Round your answer to five decimal places.


0.00020

0.00139

0.14286

0.00781

The table lists the smoking habits of a group of college students.



sex Non-smoke Regular smoke Heavy smoke total
man 135 34 5 174
woman 187 21 10 218
total 322 55 15 392

If a student is chosen at random, find the probability of getting someone who is a man or a non-smoker. Round your answer to three decimal places.

0.921

0.945

0.860

0.941

To calculate the mean and standard deviation of the number of adults over 50 who wear glasses in a random sample, we need to use the information given in the survey.

Given:
- Survey result: 72% of all adults over 50 wear glasses for driving.
- Sample size: 100 adults over 50 in the random sample.

To find the mean, we need to calculate the expected number of adults who wear glasses in the sample. We can do this by multiplying the survey result by the sample size.

Mean = Sample size * Survey result

Mean = 100 * (72 / 100) = 72 adults

The mean number of adults over 50 who wear glasses in the random sample is 72.

To calculate the standard deviation, we need to use the formula for a binomial distribution. In this case, we assume the adults wearing glasses as a random variable in a binomial distribution, with a success probability of 72% and a sample size of 100.

Standard Deviation = sqrt(N * p * (1 - p))

Standard Deviation = sqrt(100 * (0.72) * (1 - 0.72))

Standard Deviation ≈ 6.48

The standard deviation of the number of adults over 50 who wear glasses in the random sample is approximately 6.48.