Respuesta :
Answer:
17,065 of those batteries can be expected to last between three years and one month and three years and seven months
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:
I will calculate the time in months. Each year has twelve months.
Mean shelf life is three years and four months, with a standard deviation of three months. So
[tex]\mu = 3*12 + 4 = 40[/tex]
[tex]\sigma = 3[/tex]
Proportion lasting between three years and one month and three years and seven months:
This is the pvalue of Z when X = 3*12 + 7 = 43 subtracted by the pvalue of Z when X = 3*12 + 1 = 37
X = 43
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{43 - 40}{3}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413.
X = 37
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{37 - 40}{3}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587.
0.8413 - 0.1587 = 0.6826
Out of 25,000 batteries:
68.26% of the batteries are expected to last between three years and one month and three years and seven months.
0.6826*25000 = 17,065
17,065 of those batteries can be expected to last between three years and one month and three years and seven months
