62 The binomial distribution 1. Determine whether X is a binomial random variabl
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Question
62 The binomial distribution 1. Determine whether X is a binomial random variable. a. A fair die is tosses 30 times. Let X be the number of times the die come up b. b. A standard dec this deck. Let X be the number of aces. k of 52 cards contains four aces. Four cards are dealt without replacement from A SRS of 50 voters is drawn from the residents in a large city. Let X be the number who plan to vote for a proposition to increase spending on public schools. c. 2. In a recent poll, 50% of adults said that they play video games. Assume that 9 adults are randomly sampled. a. Compute the probability that exactly four of them play video games. b. Compute the probability that less than four of them play video games c. Compute the probability that at least four of them play video games. d. For a group of 9 adults, find the mean number of adults who play video games. For a group of 9 adults, find the standard deviation of the number of adults who play video games e. f. Would it be considered unusual if 8 out of 9 adults play video games. Explain.Explanation / Answer
Q2.
Binomial Experiment-
A binomial experiment (also known as a Bernoulli trial) is a statistical experiment that has the following properties:
1.The experiment consists of n repeated trials.
2.Each trial can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure.
3.The probability of success, denoted by P, is the same on every trial.
4.The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials.
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
a.
P( X = 4 ) = ( 9 4 ) * ( 0.5^4) * ( 1 - 0.5 )^5
= 0.246094
b.
P( X < 4) = P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 9 3 ) * 0.5^3 * ( 1- 0.5 ) ^6 + ( 9 2 ) * 0.5^2 * ( 1- 0.5 ) ^7 + ( 9 1 ) * 0.5^1 * ( 1- 0.5 ) ^8 + ( 9 0 ) * 0.5^0 * ( 1- 0.5 ) ^9
= 0.253906
c.
P( X < 4) = P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 9 3 ) * 0.5^3 * ( 1- 0.5 ) ^6 + ( 9 2 ) * 0.5^2 * ( 1- 0.5 ) ^7 + ( 9 1 ) * 0.5^1 * ( 1- 0.5 ) ^8 + ( 9 0 ) * 0.5^0 * ( 1- 0.5 ) ^9
= 0.253906
P( X > = 4 ) = 1 - P( X < 4) = 0.746094
d.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 9 * 0.5
= 4.5
e.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 9 * 0.5 * 0.5
= 2.25
standard deviation = sqrt( variance ) = sqrt(2.25)
=1.5
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