Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 90, \sigma = 10[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.675 = \frac{X - 90}{10}[/tex]
[tex]X - 90 = -0.675*10[/tex]
[tex]X = 83.25[/tex]
75th percentile:
X when Z has a pvalue of 0.75. So X when Z = 0.675
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.675 = \frac{X - 90}{10}[/tex]
[tex]X - 90 = 0.675*10[/tex]
[tex]X = 96.75[/tex]
The interquartile range is between 83.25 hours and 96.75 hours
