Question

In: Economics

Problem 3 Suppose that the distance of fly balls hit to the outfield (in baseball) is...

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 .


  1. (.3)   X∼    (pick one) BENU   (    ,    ) .

  2. (.35) For a random fly ball, what is the probability that this ball traveled fewer than   220   feet?   P(X<220)=    .

  3. (.35) The   80th   percentile of the distribution of fly balls is given by   P(X<  )=0.80 .

Solutions

Expert Solution

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”.


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