Question

week 7 worksheet

1. determine in which of the following situations a binomial distribution can be applied. if so, find the mean and standard deviation of x. if not, state which of the conditions to satisfy the binomial distribution requirements has been violated. round to 2 decimals.

(a) linda is interested in toilet paper pulling preferences. she takes a simple random sample of 5 people and asks each whether they always pull from the top or not. the probability that a person pulls from the top is 0.53, and x = the number of people who pull from the top.
n = 5 p = 0.53
μ = 5×0.53 = 2.65
standard deviation
σ = √(npq)
1 - p = 1 - 0.53 = 0.47 = 1.12
(b) i roll a fair, 6 - sided die until i get a two. x is the number of rolls it takes before i obtain a roll of two.
not a binomial situation because the number of trials is not fixed.
(c) you have a bag containing 4 red chips and 6 white chips and you draw 4 chips. let random variable y be the number of red chips drawn from the bag out of 4 draws without replacement.
4 red chips
6 white chips
the draws are made without replacement
so the probability will change with each draw

1. from the information in 3(a), answer the following: (round to 2 decimals.)

(a) find the probability that the number of people who pull from the top is:
i. five: p(x ) =
ii. at least three: p(x ) =
iii. between 1 and 3, not including 3: p( x ) =
iv. less than 2: p(x ) =

Explanation:

Step1: Recall binomial probability formula

The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. In part 3(a), $n = 5$ and $p=0.53$.

Step2: Calculate $P(X = 5)$

$C(5,5)=\frac{5!}{5!(5 - 5)!}=1$. Then $P(X = 5)=C(5,5)\times(0.53)^{5}\times(1 - 0.53)^{5 - 5}=1\times(0.53)^{5}\times1\approx0.04$.

Step3: Calculate $P(X\geq3)$

$P(X\geq3)=P(X = 3)+P(X = 4)+P(X = 5)$.
$C(5,3)=\frac{5!}{3!(5 - 3)!}=\frac{5\times4}{2\times1}=10$, $P(X = 3)=C(5,3)\times(0.53)^{3}\times(0.47)^{2}=10\times0.53^{3}\times0.47^{2}\approx0.34$.
$C(5,4)=\frac{5!}{4!(5 - 4)!}=5$, $P(X = 4)=C(5,4)\times(0.53)^{4}\times(0.47)^{1}=5\times0.53^{4}\times0.47\approx0.20$.
$P(X = 5)\approx0.04$. So $P(X\geq3)\approx0.34 + 0.20+0.04=0.58$.

Step4: Calculate $P(1\lt X\lt3)$

$P(1\lt X\lt3)=P(X = 2)$.
$C(5,2)=\frac{5!}{2!(5 - 2)!}=\frac{5\times4}{2\times1}=10$, $P(X = 2)=C(5,2)\times(0.53)^{2}\times(0.47)^{3}=10\times0.53^{2}\times0.47^{3}\approx0.30$.

Step5: Calculate $P(X\lt2)$

$P(X\lt2)=P(X = 0)+P(X = 1)$.
$C(5,0)=\frac{5!}{0!(5 - 0)!}=1$, $P(X = 0)=C(5,0)\times(0.53)^{0}\times(0.47)^{5}=1\times1\times0.47^{5}\approx0.02$.
$C(5,1)=\frac{5!}{1!(5 - 1)!}=5$, $P(X = 1)=C(5,1)\times(0.53)^{1}\times(0.47)^{4}=5\times0.53\times0.47^{4}\approx0.13$.
$P(X\lt2)\approx0.02 + 0.13=0.15$.

Answer:

i. $P(X = 5)\approx0.04$
ii. $P(X\geq3)\approx0.58$
iii. $P(1\lt X\lt3)\approx0.30$
iv. $P(X\lt2)\approx0.15$