2 2/3 +2 2/3 + 2 2/3 +2 2/3
2 answers:
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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 and standard deviation
, the zscore of a measure X is given by:
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
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
has a pvalue of 0.8413.
X = 37
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