Question

Suppose a geyser has a mean time between eruptions of 62 minutes. Let the interval of time between the eruptions be normally distributed with standard deviation 24 minutes. Complete parts (a) through (e) below. (a) What is the probability that a randomly selected time interval between eruptions is longer than 72 minutes? The probability that a randomly selected time interval is longer than 72 minutes is approximately 0.3372. (Round to four decimal places as needed.) (b) What is the probability that a random sample of 9 time intervals between eruptions has a mean longer than 72 minutes? The probability that the mean of a random sample of 9 time intervals is more than 72 minutes is approximately 0.1056 (Round to four decimal places as needed.) (c) What is the probability that a random sample of 24 time intervals between eruptions has a mean longer than 72 minutes? The probability that the mean of a random sample of 24 time intervals is more than 72 minutes is approximately (Round to four decimal places as needed.)

Answer 1

şolution : Given that 7 Suppose a geyser has a mean time between eruptions of 62 minutes. Let the interval of time between the eruptions be normally distributed with a standard deviation - of 24 minutes. (Q): From the given data, we have mean M = 62 and x =12 Standard deviation 5 = 24 minutes. 10 x-l Za o 72-62. 24 24 2501416 20:42 ii lze 0.42 i P(z>0.42) - Axea to the right of 0.42 az p(x772) - P(Z >0.42) 31-P (2 20:42)

☺ from the z- table, pizc 0:42) = 0.6628 = 06628. -1-06628 = 0.3372 08 33172 0%; . 3372 08 33,720% 1X > 0.3372 0 0142 (b): P(X>72) .. we have M = 62 ; a= 24 and 2- Me Ž - ling ZE n=9.

72-62 ZE- 4 21:26 ... 731-25 p( 878) = P(Z > 1.26) :-1-p (Z < 1.25) = [ - 0.8944 0.1056 P(X>72) = 0.1056 The probability that the mean of a random Sample of 9 time intervals between eruptions by ons more than 72 minutes nos approximately 10.10567

to the right of 1.25 Plz>1.95) - Area 0.1056 Ore 1.256 Z 12 (C) & p (X>72) .. we have l= 62, 0= 2t, and n=84 AŽ - Me 2 =

z =- 94 1024 24 2 - 0.0207 204 → z = 2.04 P (X>72) = P(Z > 2104) El-P(222-04) -1-0.9793 = 0.0207 1. The probability that the mean of a random Sample of 24 time intervals is more than 7a minutes is pxobability approximately 0.0207.1 -