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What proportion of cars with a mean fuel usage of 57 miles per gallon and a standard deviation of 3.5 mpg and the mpg is approximately normally distributed get over 60 mpg?
Wiki User
∙ 7y ago
z = (x - mean_x)/standard_deviation
→ z = (60 - 57)/3.5 = 0.88
Looking up z = 0.88 in normal tables gives a value of 0.3106 which is the probability that the value lies between 57 and 60, so the probability of it being greater than 60 is 0.5 - 0.3106 = 0.1894 = 18.94 %
The probability of getting greater than 60 mpg is approx 19%
Wiki User
∙ 7y ago
Wiki User
∙ 7y agoApprox 19.5695%.
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