Question

sample data). One eruption had a duration time of 110 sec.Eruptions of the Old Faithful geyser have duration times with a mean of 245.0 sec and a standard deviation of 36.4 sec (based on sample data). One eruption had a duration time of 110 sec.
a. What is the difference between a duration time of 110 sec and the mean? Answer
135

b. How many standard deviations is that (the difference found in part (a))? Answer

c. Convert the duration time of 110 sec to a z score. Answer

d. If we consider "usual" duration times to be those that convert to z scores between -2 and 2, is a duration time of 110 sec usual or unusual?

Answer 1

Answer:

a) 135 seconds

b) 3.71 standard deviations below the mean

c) Z = -3.71

d) Unusual

Step-by-step explanation:

Z-score:

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

$\mu = 245, \sigma = 36.4$

a. What is the difference between a duration time of 110 sec and the mean?

Duration of 110 seconds.

Mean of 245

245 - 110 = 135 seconds

b. How many standard deviations is that (the difference found in part (a))?

This is |Z|

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{110 - 245}{36.4}$

$Z = -3.71$

$|Z| = 3.71$

3.71 standard deviations below the mean

c. Convert the duration time of 110 sec to a z score

From b, Z = -3.71

d. If we consider "usual" duration times to be those that convert to z scores between -2 and 2, is a duration time of 110 sec usual or unusual?

Z is not in the interval of -2 and 2, so a duration time of 110 sec is unusual