12. The average time a person spends in each visit to an online social networkin
ID: 3061244 • Letter: 1
Question
12. The average time a person spends in each visit to an online social networking service is 62 minutes. The standard deviation is 12 minutes. If a visitor is selected at random, find the probability that he or she will spend the time shown on the networking ser- vice. Assume the times are normally distributed. (a) At least 180 minutes (b) At least 50 minutes 13. The average amount of snow per season in Trafford is 44 inches. The standard deviation is 6 inches. Find the probability that next year Trafford will receive the given amount of snowfall. Assume the random variable is normally distributed. (a) At most 50 inches of snow (b) At least 53 inches of snow 14. The average waiting time for a drive-in window at a local bank is 9.2 minutes, with a standard deviation of 2.6 minutes. When a customer arrives at the bank, find the probability that the customer will have to wait the given time. Assume the random variable is normally distributed. (a) Between 5 and 10 minutes (b) Less than 6 minutes or more than 9 minutes 15. The average time it takes college freshmen to complete the Mason Basic Reasoning Test is 24.6 minutes. The standard deviation is 5.8 minutes. Find these prob- abilities. Assume the random variable is normally distributed (a) It will take a student between 15 and 30 minutes to complete the test. (b) It will take a student less than 18 minutes or more than 28 minutes to complete the test.Explanation / Answer
12) a) P(X > 180) = P((X - mean)/sd > (180 - mean)/sd)
= P(Z > (180 - 62)/12)
= P(Z > 9.83)
= 1 - P(Z < 9.83)
= 1 - 1 = 0
b) P(X > 50) = P((X - mean)/sd > (500 - mean)/sd)
= P(Z > (50 - 62)/12)
= P(Z > -1 )
= 1 - P(Z < -1)
= 1 - 0.1587 = 0.8413
13) a) P(X < 50) = P((X - mean)/sd < (50 - mean)/sd)
= P(Z < (50 - 44)/6)
= P(Z < 1)
= 0.8413
b) P(X > 53) = P((X - mean)/sd > (53 - mean)/sd)
= P(Z > (53 - 44)/6)
= P(Z > 1.5)
= 1 - P(Z < 1.5)
= 1 - 0.9332 = 0.0668
14) P(5 < X < 10)
= P((5 - mean)/sd < (X - mean)/sd < (10 - mean)/sd)
= P((5 - 9.2)/2.6 < Z < (10 - 9.2)/2.6)
= P(-1.62 < Z < 0.31)
= P(Z < 0.31) - P(Z < -1.62)
= 0.6217 - 0.0526 = 0.5691
b) P(X < 6)
= P((X - mean)/sd < (6 - mean)/sd)
= P(Z < (6 - 9.2)/2.6)
= P(Z < -1.23)
= 0.1093
P(X > 9)
= P((X - mean)/sd > (9 - mean)/sd)
= P(Z > (9 - 9.2)/2.6)
= P(Z > -0.08)
= 1 - P(Z < -0.08)
= 1 - 0.4681
= 0.5319
so the required probability = 0.5319 - 0.1093 = 0.4226
15)a) P(15 < X < 30)
= P((15 - x)/sd < (X - mean)/sd < (30 - mean)/sd)
= P((15 - 24.6)/5.8 < Z < (30 - 24.6)/5.8)
= P(0.42 < Z < 0.93)
= P(Z < 0.93) - P(Z < 0.42)
= 0.8238 - 0.6628 = 0.161
b) P(X < 18)
= P((X - mean)/sd < (18 - mean)/sd)
= P(Z < (18 - 24.6)/5.8 )
= P(Z < -1.14)
= 0.1271
P(X > 18)
= P((X - mean)/sd > (28 - mean)/sd)
= P(Z > (28 - 24.6)/5.8)
= P(Z > 0.59)
= 1 - P(Z < 0.59)
= 1 - 0.7224
= 0.2776
So the required probability = 0.2776 - 0.1271 = 0.1505
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