Q3 Help pls.. urgent!
1 answer:
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Answer:
a) 0.1587 = 15.87% probability that a battery will break down during the warranty period.
b) The expected profit of the auto store on a battery is of $15.239.
c) The expected monthly profit on batteries if the auto store sells an average of 500 batteries a month is of $7,619.50.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
The life of batteries is known to be normally distributed with a mean and a standard deviation of 40 and 16 months, respectively.
This means that
a. What is the probability that a battery will break down during the warranty period?
Two year warranty, that is, 24 months. This probability is the pvalue of Z when X = 24. So
has a pvalue of 0.1587.
0.1587 = 15.87% probability that a battery will break down during the warranty period.
b. What is the expected profit of the auto store on a battery?
1 - 0.1587 = 0.8413 probability of a profit of $20.
0.1587 probability of a loss of $10. So
The expected profit of the auto store on a battery is of $15.239.
c. What is the expected monthly profit on batteries if the auto store sells an average of 500 batteries a month?
Multiplying the average for a battery by 500. So
15.239*500 = $7,619.50
The expected monthly profit on batteries if the auto store sells an average of 500 batteries a month is of $7,619.50.