Question

1. a golfer has a 40% chance of making an 8 foot putt. x = # of putt attempts before the golfer makes the putt.

a. what is the expected value for x in this problem?
b. p(x < 3)
c. p(x > 5)

Explanation:

Step1: Identify the distribution

This is a geometric distribution. In a geometric distribution, the probability of success \(p = 0.4\), and the probability of failure \(q=1 - p=0.6\). The formula for the expected value of a geometric distribution \(E(X)=\frac{q}{p}\).

Step2: Calculate the expected value for part A

Substitute \(q = 0.6\) and \(p = 0.4\) into the formula \(E(X)=\frac{q}{p}\).
\(E(X)=\frac{0.6}{0.4}=\frac{6}{4} = 1.5\)

Step3: Calculate \(P(X < 3)\) for part B

For a geometric distribution, \(P(X = k)=q^{k}p\).
\(P(X<3)=P(X = 0)+P(X = 1)+P(X = 2)\)
\(P(X = 0)=p = 0.4\)
\(P(X = 1)=q\times p=0.6\times0.4 = 0.24\)
\(P(X = 2)=q^{2}\times p=(0.6)^{2}\times0.4=0.144\)
\(P(X<3)=0.4 + 0.24+0.144=0.784\)

Step4: Calculate \(P(X>5)\) for part C

Use the formula \(P(X>n)=q^{n + 1}\)
Here \(n = 5\), so \(P(X>5)=q^{6}\)
\(P(X>5)=(0.6)^{6}=0.046656\)

Answer:

A. \(1.5\)
B. \(0.784\)
C. \(0.046656\)