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Hello, please help me how to solve problems in details for these questions please: 1.Suppose thatis normally distributed with mean 85 and standard deviation 12.

Hello, please help me how to solve problems in details for these questions please:

1.Suppose thatis normally distributed with mean 85 and standard deviation 12.

A. What is the probability thatis greater than 103.72?

Probability =

B. What value ofdoes only the top 20% exceed?

X=

2.A new car that is a gas- and electric-powered hybrid has recently hit the market. The distance travelled on 1 gallon of fuel is normally distributed with a mean of 50 miles and a standard deviation of 6 miles. Find the probability of the following events:

A. The car travels more than 53 miles per gallon.

Probability =

B. The car travels less than 43 miles per gallon.

Probability =

C. The car travels between 43 and 57 miles per gallon.

Probability =

3.Although in general you cannot know the sampling distribution of the sample mean exactly, by what distribution can you often approximate it?

A. binomial distribution

B. normal distribution

C. uniform distribution

D. none of the above

4.For each problem, select the best response:

(a) The scores of adults on an IQ test is approximately Normal with mean 100 and standard deviation 15. Corinne scores 118 on such test. Her z-score is about

A. 0.67

B. 1.2

C. 18

D. 7.87

E. None of the above.

(b) Which of these variables is least likely to have a Normal distribution?

A. Lengths of 50 newly hatched pythons.

B. Heights of 100 white pine trees in a forest.

C. Income per person for 150 different countries.

D. None of the above.

(c) To completely specify the shape of a Normal distribution, you must give

A. the median and the standard deviation.

B. the mean and the median.

C. the mean and the standard deviation.

D. the five-number summary.

E. All of the above.

5.The natural remedy echinacea is reputed to boost the immune system, which will reduce flu and colds. A 6-month study was undertaken to determine whether the remedy works. From this study, the following probability distribution of the number of respiratory infections per year (X) for echinacea users was produced:

X 0 1 2 3 4

P(X) 0.332 0.33 0.205 0.074 0.059

Find the following probabilities:

A. An echinacea user has more than one infection per year

B. An echinacea user has no infections per year

C. An echinacea user has between one and three (inclusive) infections per year

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