In the probability distribution to the​ right, the random variable X represents the number of hits a baseball player obtained in a game over the course of a season. Complete parts​ (a) through​ (f) below. x ​P(x) 0 0.1685 1 0.3358 2 0.2828 3 0.1501 4 0.0374 5 0.0254 ​

(a) Verify that this is a discrete probability distribution. This is a discrete probability distribution because all of the probabilities are at least one of the probabilities is all of the probabilities are between 0 and 1​, ​inclusive, and the sum mean sum product of the probabilities is 1. ​(Type whole numbers. Use ascending​ order.)

​(b) Draw a graph of the probability distribution. Describe the shape of the distribution. Graph the probability distribution. Choose the correct graph below. A. 0 1 2 3 4 5 0 0.1 0.2 0.3 0.4 Number of Hits Probability The graph of a probability distribution has a horizontal x-axis labeled "Number of Hits" from 0 to 5 in intervals of 1 and a vertical y-axis labeled "Probability" from 0 to 0.4 in intervals of 0.05. Vertical line segments are centered on each of the horizontal axis tick marks. The approximate heights of the vertical line segments are as follows, with the horizontal coordinate listed first and the line height listed second: 0, 0.15; 1, 0.04; 2, 0.03; 3, 0.17; 4, 0.34; 5, 0.28. B. 0 1 2 3 4 5 0 0.1 0.2 0.3 0.4 Number of Hits Probability The graph of a probability distribution has a horizontal x-axis labeled "Number of Hits" from 0 to 5 in intervals of 1 and a vertical y-axis labeled "Probability" from 0 to 0.4 in intervals of 0.05. Vertical line segments are centered on each of the horizontal axis tick marks. The approximate heights of the vertical line segments are as follows, with the horizontal coordinate listed first and the line height listed second: 0, 0.34; 1, 0.15; 2, 0.03; 3, 0.17; 4, 0.28; 5, 0.04. C. 0 1 2 3 4 5 0 0.1 0.2 0.3 0.4 Number of Hits Probability The graph of a probability distribution has a horizontal x-axis labeled "Number of Hits" from 0 to 5 in intervals of 1 and a vertical y-axis labeled "Probability" from 0 to 0.4 in intervals of 0.05. Vertical line segments are centered on each of the horizontal axis tick marks. The approximate heights of the vertical line segments are as follows, with the horizontal coordinate listed first and the line height listed second: 0, 0.03; 1, 0.04; 2, 0.15; 3, 0.28; 4, 0.34; 5, 0.17. D. 0 1 2 3 4 5 0 0.1 0.2 0.3 0.4 Number of Hits Probability The graph of a probability distribution has a horizontal x-axis labeled "Number of Hits" from 0 to 5 in intervals of 1 and a vertical y-axis labeled "Probability" from 0 to 0.4 in intervals of 0.05. Vertical line segments are centered on each of the horizontal axis tick marks. The approximate heights of the vertical line segments are as follows, with the horizontal coordinate listed first and the line height listed second: 0, 0.17; 1, 0.34; 2, 0.28; 3, 0.15; 4, 0.04; 5, 0.03. Describe the shape of the distribution. The distribution has one mode has one mode is multimodal is uniform is bimodal and is skewed right. roughly symmetric. skewed right. skewed left.

​(c) Compute and interpret the mean of the random variable X. mu Subscript xequals 0.1666 hits ​(Type an integer or a decimal. Do not​ round.) Which of the following interpretations of the mean is​ correct? A. In any number of​ games, one would expect the mean number of hits per game to be the mean of the random variable. B. Over the course of many​ games, one would expect the mean number of hits per game to be the mean of the random variable. C. The observed number of hits per game will be less than the mean number of hits per game for most games. D. The observed number of hits per game will be equal to the mean number of hits per game for most games. ​

Need help with (c) through (f) please!

(d) Compute the standard deviation of the random variable X. sigma Subscript xequals nothing hits ​(Round to three decimal places as​ needed.)

​(e) What is the probability that in a randomly selected​ game, the player got 2​ hits? nothing ​(Type an integer or a decimal. Do not​ round.)

​(f) What is the probability that in a randomly selected​ game, the player got more than 1​ hit? nothing ​(Type an integer or a decimal. Do not​ round.)

Suppose that Motorola uses the normal distribution to determine the probability of defects and the number of defects in a particular production process. Assume that the production process manufactures items with a mean weight of 10 ounces. Calculate the probability of a defect and the suspected number of defects for a 1,000-unit production run in the following situations. (a) The process standard deviation is 0.27, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects. If required, round your answer to four decimal places. (b) Through process design improvements, the process standard deviation can be reduced to 0.09. Assume that the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects. If required, round your answer to four decimal places.

Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of  ounces.

a. The process standard deviation is , and the process control is set at plus or minus  standard deviation . Units with weights less than  or greater than  ounces will be classified as defects. What is the probability of a defect (to 4 decimals)?

In a production run of  parts, how many defects would be found (round to the nearest whole number)?

b. Through process design improvements, the process standard deviation can be reduced to . Assume the process control remains the same, with weights less than  or greater than  ounces being classified as defects. What is the probability of a defect (round to 4 decimals; if necessary)?

In a production run of  parts, how many defects would be found (to the nearest whole number)?

c. What is the advantage of reducing process variation, thereby causing a problem limits to be at a greater number of standard deviations from the mean?

- Select your answer -It can substantially reduce the number of defectsIt may slightly reduce the number of defectsIt has no effect on the number of defectsItem 5

Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 11 ounces. Use Table 1 in Appendix B.

a. The process standard deviation is 0.1, and the process control is set at plus or minus 2 standard deviations. Units with weights less than 10.8 or greater than 11.2 ounces will be classified as defects. What is the probability of a defect (to 4 decimals)?

In a production run of 1000 parts, how many defects would be found (to the nearest whole number)?

b. Through process design improvements, the process standard deviation can be reduced to 0.08. Assume the process control remains the same, with weights less than 10.8 or greater than 11.2 ounces being classified as defects. What is the probability of a defect (to 4 decimals)?

In a production run of 1000 parts, how many defects would be found (to the nearest whole number)?

c. What is the advantage of reducing process variation?

Suppose that Motorola uses the normal distribution to determine the probability of defects and the number of defects in a particular production process. Assume that the production process manufactures items with a mean weight of 10 ounces. Calculate the probability of a defect and the suspected number of defects for a 1,000-unit production run in the following situations. (a) The process standard deviation is 0.30, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects. If required, round your answer to four decimal places. (b) Through process design improvements, the process standard deviation can be reduced to 0.10. Assume that the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects. If required, round your answer to four decimal places. (c) What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean? Reducing the proces standard deviation causes a in the number of defects.

Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 10 ounces.

(a)

The process standard deviation is 0.21, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.79 or greater than 10.21 ounces will be classified as defects. (Round your answer to the nearest integer.)

Calculate the probability of a defect. (Round your answer to four decimal places.)

Calculate the expected number of defects for a 1,000-unit production run. (Round your answer to the nearest integer.)

defects

(b)

Through process design improvements, the process standard deviation can be reduced to 0.07. Assume the process control remains the same, with weights less than 9.79 or greater than 10.21 ounces being classified as defects.

Calculate the probability of a defect. (Round your answer to four decimal places.)

Calculate the expected number of defects for a 1,000-unit production run. (Round your answer to the nearest integer.)

defects

(c)

What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean?

Reducing the process standard deviation causes no change in the number of defects.Reducing the process standard deviation causes a substantial increase in the number of defects.     Reducing the process standard deviation causes a substantial reduction in the number of defects.

In the binomial probability distribution, let the number of trials be n = 4, and let the probability of success be p = 0.3310. Use a calculator to compute the following.

(a) The probability of three successes. (Round your answer to three decimal places.)


(b) The probability of four successes. (Round your answer to three decimal places.)


(c) The probability of three or four successes. (Round your answer to three decimal places.)

A valuable object worth $10,000 has the probability 1/4 of being damaged accidentally. The damaged object will be worth $6,400. The owner’s utility function is u(w) = w^(1/2).

(a) The price of an insurance against losing the $3,600 is $900. Will the owner be willing to buy the insurance?

(b) What is the highest price that the owner would be willing to pay to insure against the $3,600 loss?