Answer:
0.7881 = 78.81% probability that a randomly chosen Ford truck outperforms a randomly chosen GM truck in terms of fuel efficiency
Step-by-step explanation:
When the distribution is normal, we use 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.
Subtraction of normal variables:
When we subtract normal variables, the mean is the subtraction of the means, and the standard deviation is the square root of the sum of the variances.
The fuel efficiency of Ford trucks is found to follow a normal distribution with mean 20 mpg and a standard deviation of 3 mpg.
This means that
The fuel efficiency of GM trucks is found to follow a normal distribution with mean 24 mpg and standard deviation of 4 mpg.
This means that
Compute the probability that a randomly chosen Ford truck outperforms a randomly chosen GM truck in terms of fuel efficiency
This is the probability that the subtraction of ford by gm is less than 0. First, we find the mean and the standard deviation.
This probability is the pvalue of Z when X = 0. So
has a pvalue of 0.7881
0.7881 = 78.81% probability that a randomly chosen Ford truck outperforms a randomly chosen GM truck in terms of fuel efficiency