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According to a certain golf​ association, the weight of the golf ball shall not be greater than 1.620 ounces​ (45.93 grams). The diameter of the ball shall not be less than 1.680 inches. The velocity of the ball shall not be greater than 250 feet per second. The golf association periodically checks the specifications of golf balls using random sampling. Five dozen of each kind are​ sampled, and if more than three do not meet size or velocity​ requirements, that kind of ball is removed from the golf​ association's approved list. Complete parts a through c below.

a. What assumptions must be made in order to use the binomial probability distribution to calculate the probability that a particular kind of golf ball will be​ removed?

A. The experiment consists of n identical trials. There are only two possible outcomes on each trial. The probability of success can change from trial to trial. The trials are dependent.
B. The experiment consists of n identical trials. There are only two possible outcomes on each trial. The probability of success remains the same from trial to trial. The trials are independent.
C. The experiment consists of n identical trials. The number of outcomes can vary. The probability of success can change. The trials are independent.

b. What information must be known in order to use the binomial probability distribution to calculate the probability that a particular kind of golf ball will be​ removed?
A. The percentage of that kind of golf ball that meets all the requirements
B. The percentage of that kind of golf ball that meets the size requirements
C. The percentage of that kind of golf ball that meets the velocity requirements
D. The percentage of that kind of golf ball that fails to meet both size and velocity requirements

c. Suppose 20​% of all balls produced by a particular manufacturer are less than 1.680 inches in​ diameter, and assume that the number of such​ balls, x, in a sample of fivefive dozen balls can be adequately characterized by a binomial probability distribution.

Find the mean of the binomial distribution. ​(Round to the nearest tenth as​ needed.)
Find the standard deviation of the binomial distribution. (Round to the nearest thousandth as​ needed.)

Respuesta :

MrRoyal

Missing Part of Question

d. Refer to part c . If x has a binomial distribution, then so does the number, y , of balls in the sample that meet the USGA’s minimum diameter. [ Note: x + y = 60 ] Describe the distribution of y . In particular, what are p, q , and n ? Also, find E(y) and the standard deviation of y .

Answer:

c. Mean = 12

Standard Deviation = 3.098

d.

n = 60

p = 0.8

q = 0.2

Mean = 48

Standard Deviation = 3.098

Step-by-step explanation:

You've answered a and b part of the question, already.

See extracts below

------------------- Answered to Question a and b begins here

"a. What assumptions must be made in order to use the binomial probability distribution to calculate the probability that a particular kind of golf ball will be removed?

A. The experiment consists of n identical trials. There are only two possible outcomes on each trial. The probability of success can change from trial to trial. The trials are dependent.

B. The experiment consists of n identical trials. There are only two possible outcomes on each trial. The probability of success remains the same from trial to trial. The trials are independent.

C. The experiment consists of n identical trials. The number of outcomes can vary. The probability of success can change. The trials are independent.

b. What information must be known in order to use the binomial probability distribution to calculate the probability that a particular kind of golf ball will be removed?

A. The percentage of that kind of golf ball that meets all the requirements

B. The percentage of that kind of golf ball that meets the size requirements

C. The percentage of that kind of golf ball that meets the velocity requirements

D. The percentage of that kind of golf ball that fails to meet both size and velocity requirements

--------------------------- Answer to Question a and b stops here

I'll continue from c and then proceed to d..

c.

Mean is calculated by np

where n = number of observation i.e. sample size = 5 dozens = 5 * 12 = 60

p = probability of success = 20% = 0.2

Mean, E(x) = 0.2 * 60 = 12

Standard Deviation is calculated by the SquareRoot of the products of mean by q (

Standard Deviation = √npq

Where q = 1 - p

q = 1 - 0.2

q = 0.8

Standard Deviation = √(12)(0.8)

Standard Deviation = √9.6

Standard Deviation = 3.098386676965933

Standard Deviation = 3.098 ----- Approximated

d.

From (c) above,

x represents the number of balls that didn't meet USGA minimum diameter

And y represents the number of balls that meet USGA minimum diameter

In total, there are 5 dozen balls

So, x + y = 5 * 12

x + y = 60

So, n = x + y = 60

p, the probability of success of y is the complement of the probability of success of x.

i.e. p of y = 1 - 0.2

p = 0.8

q = 1 - 0.8 = 0.2

Mean is calculated as

E(Y) = np = 60 * 0.8 = 48

Standard Deviation = √npq

Standard Deviation = √(60 * 0.8 * 0.2)

Standard Deviation = √9.6

Standard Deviation = 3.098386676965933

Standard Deviation = 3.098 ----- Approximated