320 Section 7.4 Problems ill be What is the probability that the sample mean w l

Question

320 Section 7.4 Problems ill be What is the probability that the sample mean w less than $7.45? c) be d) What is the probability that the sample mean will more than $7.60? 7.13 In an effort to the shorten the length of games, Major League Baseball instituted a new rule in 2007 that stated that pitchers had 12 seconds to pitch the ball to a batter Unfortunately, the rule has not been enforced by umpires very often. The Wall Street Journal recently performed a study on several pitchers to determine SLOTH, or Seconds to Launch One Throw Home. Of 30 pitchers measured, Josh Beckett of the Boston Red Sox had the highest SLOTH average, 14.88 seconds. Assume Beckett's SLOTH values per pitch follow the normal probability distribution with a standard deviation of 2.5 seconds. a) What is the probability that the average of a random sample of 10 pitches from Josh Beckett will have a mean less than 14 seconds? b) What is the probability that the average of a random sample of 20 pitches from Josh Beckett will have a mean less than 14 seconds? c) What is the probability that the average of a random sample of 30 pitches from Josh Beckett will have a mean less than 14 seconds? d) Explain the difference in these probabilities.

Explanation / Answer

Question 7.13

Mean of Josh' SLOTH average = 14.88 seconds

Standard deviation of Josh's SLOTH average = 2.5 seconds

(a) standard error of the sample mean of sample size 10 = /sqrt(n) = 2.5/ sqrt(10) = 0.79

So, Pr( x < 14 ; 14.88; 0.79) = ?

Z = (14 - 14.88)/0.79 = -1.11

Pr( x < 14 ; 14.88; 0.79) = Pr(Z < -1.11) = 0.1335

(b) Now sample size = 20

standard error of the sample mean of sample size 20 = /sqrt(n) = 2.5/ sqrt(20) = 0.559

So, Pr( x < 14 ; 14.88; 0.559) = ?

Z = (14 - 14.88)/0.559 = -1.5742

Pr( x < 14 ; 14.88; 0.559) = Pr(Z < -1.11) = 0.0577

(c) Now sample size = 30

standard error of the sample mean of sample size 30 = /sqrt(n) = 2.5/ sqrt(30) = 0.4564

So, Pr( x < 14 ; 14.88; 0.4564) = ?

Z = (14 - 14.88)/0.4564 = -1.9281

Pr( x < 14 ; 14.88; 0.559) = Pr(Z < -1.9281) = 0.0269

(d) These probabilities are different it is vecause the sample size has been increasing that will reduce the standard deviation of the difference in mean.

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