Question

8. Suppose weights of cars in America are normally distributed with a mean of 3550lbs and a standard deviation of 870lbs.

1. What is the probability that a car weighs between 3000 and 4000 lbs?

2. What would be the weight of a car in the 84th percentile of this distribution?

Answer 1

This is a normal distribution question with

P(3000.0 < x < 4000.0)=?

This implies that

P(3000.0 < x < 4000.0) = P(-0.6322 < z < 0.5172) = P(Z < 0.5172) - P(Z < -0.6322)

P(3000.0 < x < 4000.0) = 0.6974917239816899 - 0.263628098662785

Given in the question

P(X < x) = 0.84

This implies that

P(Z < 0.994457883209753) = 0.84

With the help of formula for z, we can say that

PS: you have to refer z score table to find the final probabilities.

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