Question

unit 4. probability, random variables, and probability distributions
hw 8 – the geometric distribution
name: lucy luo

1. for the following situations, decide if it is a binomial setting, a geometric setting, or neither. explain your answer.

a) you keep drawing cards out a deck, without replacement, until an ace is drawn.
two possible outcomes : ace/not ace
fixed probability :
independent trials : x
first success : v
neither
b) you roll a dice 20 times and record the number of sixes you have rolled.
two outcomes : six/not six
fixed probability
independent trials : v
c) crest claims that 40% of americans use their toothpaste. you take a random sample of 50 americans and count how many use crest toothpaste.
two outcomes : use toothpaste/not use
fixed probability :.40
d) you flip a coin until you get tails.
two outcomes : head/tail
fixed probability :.50
independent trials : v
e) 5% of the tomatoes at a farmer’s market have imperfections on them. you randomly choose one tomato at a time until you find one with an imperfection.
two outcomes : imperfection/perfection
fixed probability :.05
independent trials : v
first success : v geometric
f) 5% of the tomatoes at the farmer’s market have imperfections on them. you randomly choose 20 tomatoes and count the number of imperfections on them.
two outcomes : imperfection/perfe
fixed probability :.05
independent trials : v

1. in a recent pew study, it was found that 28% of all homes have at least one dog. let x = the number of houses you look at before finding a house with a dog. find and interpret the following probabilities.

a) p(x = 3)
b) p(x < 5)
c) p(x > 4)
d) e(x)

Explanation:

Step1: Identify the distribution

Since we are looking for the number of trials before the first success (finding a house with a dog), this is a geometric distribution. The probability of success \(p = 0.28\), and the probability of failure \(q=1 - p= 0.72\)

Step2: Calculate \(P(X = 3)\)

The formula for the geometric distribution is \(P(X = k)=q^{k - 1}p\). For \(k = 3\), we have:

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Step3: Calculate \(P(X\lt5)\)

\(P(X\lt5)=P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)\)

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Step4: Calculate \(P(X\gt4)\)

We know that \(P(X\gt4)=1 - P(X\leq4)\). Since \(P(X\leq4)=P(X\lt5)\), so \(P(X\gt4)=1 - 0.731262 = 0.268738\)

Step5: Calculate \(E(X)\)

The expected value of a geometric distribution is \(E(X)=\frac{1}{p}\). Substituting \(p = 0.28\), we get \(E(X)=\frac{1}{0.28}\approx3.57\)

Answer:

a) \(P(X = 3)=0.145\) (rounded to three decimal places)
b) \(P(X\lt5)=0.731\) (rounded to three decimal places)
c) \(P(X\gt4)=0.269\) (rounded to three decimal places)
d) \(E(X)\approx3.57\)