Question

The length of time, M hours that a phone battery will work before it needs recharging is normally distributed with a mean of 90 hours, and a standard deviation of 10 hours. Find the interquartile range for this distribution.

Answer 1

Answer:

The interquartile range is between 83.25 hours and 96.75 hours

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

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 problem, we have that:

$\mu = 90, \sigma = 10$

Find the interquartile range for this distribution.

That is, the middle 50%, from the 25th to the 75th percentile.

25th percentile:

X when Z has a pvalue of 0.25. So X when Z = -0.675

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

$-0.675 = \frac{X - 90}{10}$

$X - 90 = -0.675*10$

$X = 83.25$

75th percentile:

X when Z has a pvalue of 0.75. So X when Z = 0.675

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

$0.675 = \frac{X - 90}{10}$

$X - 90 = 0.675*10$

$X = 96.75$

The interquartile range is between 83.25 hours and 96.75 hours