Respuesta :
Answer:
92.10% probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long.
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 = 30.05, \sigma = 0.2[/tex]
What is the probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long?
This is the pvalue of Z when X = 30.5 subtracted by the pvalue of Z when X = 29.75
X = 30.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{30.5 - 30.05}{0.2}[/tex]
[tex]Z = 2.25[/tex]
[tex]Z = 2.25[/tex] has a pvalue of 0.9878
X = 29.75
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{29.75 - 30.05}{0.2}[/tex]
[tex]Z = -1.5[/tex]
[tex]Z = -1.5[/tex] has a pvalue of 0.0668
So there is a 0.9878 - 0.0668 = 0.9210 = 92.10% probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long.
