Respuesta :
Answer:
a) The random variable is the average life of a refrigerator.
b) 0.1151 = 11.51% probability that someone will keep a refrigerator fewer than 11 years before replacement.
c) 0.0548 = 5.48% probability that someone will keep a refrigerator more than 18 years before replacement
d) 0.6472 = 64.72% probability that someone will replace a refrigerator between 8 and 15 years
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.
Consumer Reports indicated that the average life of a refrigerator before replacement is 14 years with a standard deviation of 2.5 years
This means that:
The random variable is the average life of a refrigerator.
[tex]\mu = 14, \sigma = 2.5[/tex]
a. State the random variable.
The random variable is the average life of a refrigerator.
b. What is the probability that someone will keep a refrigerator fewer than 11 years before replacement?
This is the pvalue of Z when X = 11. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{11 - 14}{2.5}[/tex]
[tex]Z = -1.2[/tex]
[tex]Z = -1.2[/tex] has a pvalue of 0.1151
0.1151 = 11.51% probability that someone will keep a refrigerator fewer than 11 years before replacement.
c. What is the probability that someone will keep a refrigerator more than 18 years before replacement?
This is 1 subtracted by the pvalue of Z when X = 18. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{18 - 14}{2.5}[/tex]
[tex]Z = 1.6[/tex]
[tex]Z = 1.6[/tex] has a pvalue of 0.9452
1 - 0.9452 = 0.0548
0.0548 = 5.48% probability that someone will keep a refrigerator more than 18 years before replacement.
d. What is the probability that someone will replace a refrigerator between 8 and 15 years?
This is the pvalue of Z when X = 15 subtracted by the pvalue of Z when X = 8.
X = 15
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{15 - 14}{2.5}[/tex]
[tex]Z = 0.4[/tex]
[tex]Z = 0.4[/tex] has a pvalue of 0.6554
X = 8
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{8 - 14}{2.5}[/tex]
[tex]Z = -2.4[/tex]
[tex]Z = -2.4[/tex] has a pvalue of 0.0082
0.6554 - 0.0082 = 0.6472
0.6472 = 64.72% probability that someone will replace a refrigerator between 8 and 15 years
