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ddaly55
28.07.2019 •
Mathematics
The miles-per-gallon obtained by the 1995 model q cars is normally distributed with a mean of 22 miles-per-gallon and a standard deviation of 5 miles-per-gallon. what is the probability that a car will get less than 21 miles-per-gallon? (4 decimal format = 0.0000)
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Ответ:
The probability that a car will get less than 21 miles-per-gallon is 0.4207.
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, we have that:
What is the probability that a car will get less than 21 miles-per-gallon?
This probability is the pvalue of Z when
. So
So the probability that a car will get less than 21 miles-per-gallon is 0.4207.
Ответ:
Ответ:
1. X= vinegar, Y= oil
2. o=1.5v