Suppose a geyser has a mean time between eruptions of 66 minutes. If the interva
ID: 3430429 • Letter: S
Question
Suppose a geyser has a mean time between eruptions of 66 minutes. If the interval of time between the eruptions is normally distributed with standard deviation 17 minutes, answer the following questions. (a) What is the probability that a randomly selected time interval between eruptions is longer than 73 minutes? The probability that a randomly selected time interval is longer than 73 minutes is approximately . (Round to four decimal places as needed.) (b) What is the probability that a random sample of 16 time intervals between eruptions has a mean longer than 73 minutes? The probability that the mean of a random sample of 16 time intervals is more than 73 minutes is approximately . (Round to four decimal places as needed.) (c) What is the probability that a random sample of 39 time intervals between eruptions has a mean longer than 73 minutes? The probability that the mean of a random sample of 39 time intervals is more than 73 minutes is approximately. (Round to four decimal places as needed.)Explanation / Answer
Mean = u = 66
SD = 17
a) greater than 73 minutes :
x = 73
z = (x - u) / SD
z = (73 - 66) / 17
z = 0.4117647058823529
Now, using this link :
https://www.easycalculation.com/statistics/p-value-for-z-score.php
Enter the z-value :
P(z > 0.4117647058823529) = 0.3403 --> FIRST ANSWER
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For a sample,
z = (x - u) / (SD/sqrt(n))
z = (73 - 66) / (17/sqrt(16))
z = 1.6470588235294118
P(z > 1.6470588235294118) = 0.0498 --> SECOND ANSWER
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For a sample,
z = (x - u) / (SD/sqrt(n))
z = (73 - 66) / (17/sqrt(39))
= 2.5714697640463993
P(z > 2.5714697640463993) = 0.0051 ---> THIRD ANSWER
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