Respuesta :
Using the normal distribution, we have that:
a) There is a 0.6915 = 69.15% chance of a pride of lions having a territory less than 74-square miles.
b) There is a 0.7888 = 78.88% chance of a pride of lions having a territory between 60 and 80 square miles.
c) X = 54.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
[tex]\mu = 70, \sigma = 8[/tex]
The chance of a pride of lions having a territory less than 74-square miles is the p-value of Z when X = 74, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (74 - 70)/8
Z = 0.5.
Z = 0.5 has a p-value of 0.6915.
Hence:
There is a 0.6915 = 69.15% chance of a pride of lions having a territory less than 74-square miles.
The change of the territory being between 60 and 80 square miles is the p-value of Z when X = 80 subtracted by the p-value of Z when X = 60, hence:
X = 80:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (80 - 70)/8
Z = 1.25
Z = 1.25 has a p-value of 0.8944.
X = 60:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (60 - 70)/8
Z = -1.25
Z = -1.25 has a p-value of 0.1056.
0.8944 - 0.1056 = 0.7888.
Hence:
There is a 0.7888 = 78.88% chance of a pride of lions having a territory between 60 and 80 square miles.
For item c, we have that X is found when Z = -2, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
-2 = (X - 70)/8
X - 70 = -2(8)
X = 54.
More can be learned about the normal distribution at https://brainly.com/question/4079902
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