Recently, the bowling scores of a certain bowler were normally distributed with
ID: 2261044 • Letter: R
Question
Recently, the bowling scores of a certain bowler were normally distributed with mean 199 and standard deviation 23. a) Find the probability that a score is from 180 to 210, and interpret your results. b) Find the probability that a score is from 155 to 170, and interpret your results. c) Find the probability that a score is greater than 195, and interpret your results. Click the icon to view the standard normal distribution table. a) The probability that a score is from 180 to 210 is Round to four decimal places as needed.) Interpret your results. Select the correct choice below and fill in the answer box to complete your choice. Round to the nearest whole number as needed.) A. About % of the bowler's scores are between 180 and 195. B. About % of the bowler's scores are between 195 and 210. C. About % of the bowler's scores are between 180 and 210. b) The probability that a score is from 155 to 170 is Round to four decimal places as needed.) Interpret your results. Select the correct choice below and fill in the answer box to complete your choice. Round to the nearest whole number as needed.) A. About % of the bowler's scores are between 155 and 195. B. About % of the bowler's scores are between 155 and 170. Click to select your answer(s).Explanation / Answer
a)
P(180 < X<210) = P(X < 210) - P(X < 180)
= P(Z < 210-199/23) - P(Z < 180-209/23)
= P(Z < 0.4782) - P(Z < -1.261)
= 0.6837 - 0.1037
= 0.58
b)
P(155 < X<170) = P(X < 170) - P(X < 155)
= P(Z < (170-199)/23) - P(Z < (155-199)/23)
= P(Z < -1.2609) - P(Z < -1.9131)
=0.1037 - 0.0279
= 0.0758
c)
P(X >195) = P(Z > (195 - 199)/23)
= P(Z > -0.1739)
= 0.5690
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