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The United States Golf Association requires that the weight of a golf ball must not exceed...

The United States Golf Association requires that the weight of a golf ball must not exceed 1.62 oz. The association periodically checks golf balls sold in the United States by sampling specific brands stocked by pro shops. Suppose that a manufacturer claims that no more than 9 percent of its brand of golf balls exceed 1.62 oz. in weight. Suppose that 24 of this manufacturer's golf balls are randomly selected, and let x denote the number of the 24 randomly selected golf balls that exceed 1.62 oz. Refer to the Binomial table given below.

  

Excel Output of the Binomial Distribution with n = 24, p = 0.09, and q = 0.91
  
Binomial distribution with n = 24 and p = 0.09

  

x P(X=x)
0 0.1040
1 0.2468
2 0.2807
3 0.2036
4 0.1057
5 0.0418
(a)

Find P(x = 0), that is, find the probability that none of the randomly selected golf balls exceeds 1.62 oz. in weight. (Use table values rounded to 4 decimal places for calculations. Round your answer to 4 decimal places.)

(b)

Find the probability that at least one of the randomly selected golf balls exceeds 1.62 oz. in weight. (Use table values rounded to 4 decimal places for calculations. Round your answer to 4 decimal places.)

(c)

Find P(x ≤ 3). (Use table values rounded to 4 decimal places for calculations. Round your answer to 4 decimal places.)

(d)

Find P(x ≥ 2). (Use table values rounded to 4 decimal places for calculations. Round your answer to 4 decimal places.)

(e)

Suppose that 2 of the 24 randomly selected golf balls are found to exceed 1.62 oz. Using your result from part d, do you believe the claim that no more than 9 percent of this brand of golf balls exceed 1.62 oz. in weight?

  (Click to select)NoYes , the probability of this result is (Click to select)largesmall if the claim is true.

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Answer #1

a)

P(X=0) =0.1040

b)

probability that at least one of the randomly selected golf balls exceeds 1.62 oz. in weight =1-P(X=0)

=1-0.1040 =0.8960

c)

P(x ≤ 3) =0.1040+0.2468+0.2807+0.2036=0.8351

d)

P(x>=2)=1-P(X=0)-P(X=1) =1-0.1040-0.2468 =0.6492

e)

Yes the probability of this result is large  if the claim is true.

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