In: Economics
Problem 3
Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 270 feet and a standard deviation of 50 feet. Let X= distance in feet for a fly ball.
In each appropriate box you are to enter either a rational number in "p/q" format or a decimal value accurate to the nearest 0.01 .
a).
Consider the given problem here “X” be a random variable denotes “distance in feet for a fly ball”. Now, “X” follows normal distribution with mean “270” and standard deviation “50”. So, if we pick any one then the probability distribution must be same, => “xi” also follows the normal distribution with mean “270” and standard deviation “50”.
c).
The probability that a ball traveled fewer than 220 feet is given by.
=> P(X < 220) = P[(X-270)/50 < (220-270)/50] = P[t < (-1)] = 1 - P[t >1] = 1 – 0.8413447 = 0.158655 = 0.16, where “t” be a standard normal variable. So, the required probability is given by “P(X < 220) = 0.16”.
e).
Let’s assume “t” be standard normal variable having mean “0” and standard deviation “1”. So, here the value of “P80” is given by, “P80 = 0.85” (from the standard normal table). Since, for “t < 0.85” the probability is approximately “0.8”. So, here the value of “X” is “x80” for which “P(x1<x80) = 0.8”.
So, “x8 = P80*50 + 270 = 0.85*50 + 270 = 312.5. So, the “80th” percentile value is given by, “312.5”.