Question 1 (1 point) Red Bull is the most popular energy drink in sales in the U

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Question 1 (1 point)

Red Bull is the most popular energy drink in sales in the United States. Red Bull GmbH (the parent company) has observed that daily sales are normally distributed with an average of 6,453,402 drinks sold with a standard deviation of 5,123.084. What is the probability that on a given day above 6,448,785 drinks are sold?

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Question 2 (1 point)

If the scores of golfers on the PGA tour are approximately normally distributed with mean 68.39 and standard deviation 1.556, what is the probability that a randomly chosen golfer's score is below 67.1 strokes?

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Question 3 (1 point)

If the scores of golfers on the PGA tour are approximately normally distributed with mean 66.77 and standard deviation 2.981, what is the probability that a randomly chosen golfer's score is above 69.1 strokes?

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Question 4 (1 point)

If the scores of golfers on the PGA tour are approximately normally distributed with mean 67.25 and standard deviation 2.745, what is the probability that a randomly chosen golfer's score is below 62.2 strokes?

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Question 5 (1 point)

Suppose that the mean and standard deviation of the number of gallons of milk sold at a local supermarket per day are 194.019 and 15.4879, respectively. Fill in the blank: the supermarket will sell greater than __________ gallons of milk on 69.85% of days. Assume the distribution is approximately normal.

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Question 6 (1 point)

Suppose that the distribution of income in a certain tax bracket is approximately normal with a mean of $24,712.02 and a standard deviation of $4,159.539. Approximately 51.53% of households had an income less than what dollar amount?

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Question 7 (1 point)

Suppose that the mean and standard deviation of the number of gallons of milk sold at a local supermarket per day are 199.403 and 10.0199, respectively. Fill in the blank: the supermarket will sell less than __________ gallons of milk on 99.11% of days. Assume the distribution is approximately normal.

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Question 8 (1 point)

Suppose that the mean and standard deviation of the number of gallons of milk sold at a local supermarket per day are 216.649 and 10.4108, respectively. Fill in the blank: the supermarket will sell greater than __________ gallons of milk on 53.43% of days. Assume the distribution is approximately normal.

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Question 9 (1 point)

Suppose that the mean and standard deviation of the scores on the mathematics portion of the SAT are 482.56 and 56.591, respectively, for a given year. 16.82% of students scored greater than what score? Assume the distribution is approximately normal.

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Question 10 (1 point)

Suppose that the lifespan of beagles is normally distributed with a mean of 10.91 and a standard deviation of 0.823. Would it be unusual to observe a beagle lifespan below 9.8?

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Question 11 (1 point)

The composite ACT score for high school students in their senior year is normally distributed with a mean of 19.79 and a standard deviation of 5.557. Would it be unusual for a randomly selected student to have a score below 4?

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Question 12 (1 point)

The composite ACT score for high school students in their senior year is normally distributed with a mean of 20.99 and a standard deviation of 5.863. Would it be unusual for a randomly selected student to have a score between 7.1 and 35.2?

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Question 13 (1 point)

The number of home runs hit for each player in Major League Baseball are approximately normally distributed with a mean of 21.58 and a standard deviation of 2.006. Would it be unusual for a randomly selected player of have hit below 18.5 homeruns in the previous season?

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Question 14 (1 point)

The number of home runs hit for each player in Major League Baseball are approximately normally distributed with a mean of 21.03 and a standard deviation of 2.887. Would it be unusual for a randomly selected player of have hit below 16.5 homeruns in the previous season?

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1) 0.1590 2) 0.8163 3) 0.1837 4) 0.8410 5) We do not have enough information to calculate the value.

Explanation / Answer

Solution:

1) There is a standard formula used in such cases

z=X - mue(average)/standard deviation

and then find the associated z value in the z table

so we have 6,448,785 - 6453402/5123.084= -0.90

checking the z value of 0.90 in z table we get 0.8163

2) solving with same z equation we get 0.8311

3) 0.7828

4)0.8117

5) 202.07, explanation : since the data is normally distributed, 68% of values is covered under 1 standard deviation which is = 202.7 or more

6)24552.45 explnation : same as above

7)175.66

8)207.52

9)428.2

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