Suppose a geyser has a mean time between eruption of 72 minu

Suppose a geyser has a mean time between eruption of 72 minutes. If the interval of time between eruptions is normally distributed with standard deviation 20 minutes, answer the following questions.

a) What is the probability that a randomly selcted interval between eruptions is longer than 81 minutes. (Round to Four Decimal Places as needed)

b) What is the probability that a random sample of 12 time intervals between eruptions have mean longer than 81 minutes?

c) What is the probability that a random sample of 20 time intervals between eruptions has a mean longer than 81 minutes?

d) What effect does increasing the sameple size have on the probability?

1) The probability increases because the variability in the sample mean increase as the sample size increases

2) The probability decreases because the variability in the sample mean increases as the sample size increases

3) The probability decreases because the variability in the sample mean decreases as the sample size increases.

4) The probability increases because the variability in the sample mean decreases as the sample size increases

e) What might you concluded if a random sample of 20 time intervals between eruptions has a mean longer than 81 minutes?

1) The population mean must be less than 72 since the probability is so low

2 The population mean is 72 minutes, and this is an example of typical sampling

3 The population mean cannot be 72, since the probability is so low

4 The population mean may be greather than 72

Solution

(a) P(X>81) = P((X-mean)/s >(81-72)/20)

=P(Z>0.45) = 0.3264 (from standard normal table)

--------------------------------------------------------------------------------------------------------------------

(b)P(xbar>81) = P((xbar-mean)/(s/vn) >(81-72)/(20/sqrt(12)))

=P(Z>1.56) = 0.0594 (from standard normal table)

--------------------------------------------------------------------------------------------------------------------

(c)P(xbar>81) = P((xbar-mean)/(s/vn) >(81-72)/(20/sqrt(20)))
=P(Z>2.01) = 0.0222 (from standard normal table)

--------------------------------------------------------------------------------------------------------------------

(d) 3) The probability decreases because the variability in the sample mean decreases as the sample size increases.

--------------------------------------------------------------------------------------------------------------------

(e)2 The population mean is 72 minutes, and this is an example of typical sampling