i) Ted\'s bowling scores are approximately normally distributed with mean 170 an
ID: 3135343 • Letter: I
Question
i) Ted's bowling scores are approximately normally distributed with mean 170 and standard deviation 13, while Pam's scores are normally distributed with mean 155 and standard deviation 12. If Ted and Pam each bowl one game, then assuming that their scores are independent random variables, approximate the probability that the total of their scores is above 340.
ii) Suppose that the average number of airline crashes in a country is 2.4 per half year.
(a) What is the probability that there will be at least 2 accidents in the next half year?
(b) What is the probability that there will be at least 4 accidents in the next year?
(c) What is the probability that there will be at most 6 accidents in the next 1.5 years?
Explanation / Answer
i) Ted's bowling scores are approximately normally distributed with mean 170 and standard deviation 13, while Pam's scores are normally distributed with mean 155 and standard deviation 12. If Ted and Pam each bowl one game, then assuming that their scores are independent random variables, approximate the probability that the total of their scores is above 340.
Total score mean =170+155 =325
Total score variance =13^2 +12^2 =313
Total sd =17.6918
Z value for 340, z=(340-325)/17.6918 =0.85
P( x >340) = P( z >0.85)
=0.1977
ii) Suppose that the average number of airline crashes in a country is 2.4 per half year.
(a) What is the probability that there will be at least 2 accidents in the next half year?
Poisson distribution used.
Mean for half year =2.4
P( x 2) =0.6916
(b) What is the probability that there will be at least 4 accidents in the next year?
Mean for one year =2.4*2=4.8
P(x 4) = 0.7058
(c) What is the probability that there will be at most 6 accidents in the next 1.5 years?
Mean for 1.5 year =2.4*3=7.2
P( x 6 ) = 0.4204
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