Respuesta :
Answer:
0.8041 = 80.41% probability that a given battery will last between 2.3 and 3.6 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.
A certain type of storage battery lasts, on average, 3.0 years with a standard deviation of 0.5 year
This means that [tex]\mu = 3, \sigma = 0.5[/tex]
What is the probability that a given battery will last between 2.3 and 3.6 years?
This is the p-value of Z when X = 3.6 subtracted by the p-value of Z when X = 2.3. So
X = 3.6
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3.6 - 3}{0.5}[/tex]
[tex]Z = 1.2[/tex]
[tex]Z = 1.2[/tex] has a p-value of 0.8849
X = 2.3
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{2.3 - 3}{0.5}[/tex]
[tex]Z = -1.4[/tex]
[tex]Z = -1.4[/tex] has a p-value of 0.0808
0.8849 - 0.0808 = 0.8041
0.8041 = 80.41% probability that a given battery will last between 2.3 and 3.6 years
