Respuesta :
Answer:
P(66.4 < X < 241.6) = 0.5039 = 50.39%.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
A distribution of values is normal with a mean of 232.4 and a standard deviation of 92.2.
This means that [tex]\mu = 232.4, \sigma = 92.2[/tex]
Find the probability that a randomly selected value is between 66.4 and 241.6.
This is the p-value of Z when X = 241.6 subtracted by the p-value of Z when X = 66.4.
X = 241.6
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{241.6 - 232.4}{92.2}[/tex]
[tex]Z = 0.1[/tex]
[tex]Z = 0.1[/tex] has a p-value of 0.5398
X = 66.4
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{66.4 - 232.4}{92.2}[/tex]
[tex]Z = -1.8[/tex]
[tex]Z = -1.8[/tex] has a p-value of 0.0359
0.5398 - 0.0359 = 0.5039
P(66.4 < X < 241.6) = 0.5039 = 50.39%.
