Question

in a certain population, 11% of people are left-handed. Suppose that you plan to randomly select 100 people and ask each person whether they are left handed. Suppose that in calculating each of the probabilities below, you use the normal distribution as an approximation to the binomial but that you fail to use a continuity correction. In which cases will you obtain an answer that is too large? A: the probability that among the 100 people, at least 12 are left-handed B: the probability that among the 100 people, more than 12 are left-handed C: the probability that among the 100 people, between 10 and 14 inclusive are left-handed Choose best response a. B only b. A only c. A and C d. B and C

Answer 1

c. A and C

Explanation:

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

`X \sim Binom(n=100, p=0.11)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

We need to check the conditions in order to use the normal approximation.

`np=100*0.11=11 > 10 \geq 10`

`n(1-p)=100*(1-0.11)=99 \geq 10`

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

`E(X)=np=100*0.11=11`

`\sigma=√(np(1-p))=√(100*0.11(1-0.11))=3.129`

Part A

We want this probability:

`P(X \geq 12) = 1-P(X<12) = 1-P(X\leq 11)`

The z score is defined as

`Z=(x-\mu)/(\sigma)`.

`P(X \geq 12) = 1-P(X<11) = 1-P(Z< (11-11)/(3.129))=1-P(Z<0)=1-0.5=0.5`

Part B

P(X>12) = 1-P(X\leq 12) = 1-P(Z< \frac{12-11}{3.129})=1-0.625=0.375[/tex]

Part C

`P(10\leq X \leq 14) = P(X<14)-P(X<10)`

The z score is defined as

`Z=(x-\mu)/(\sigma)`.

`P(10 \leq X \leq 14) =P(Z< (14-11)/(3.129)) -P(Z< (10-11)/(3.129))=P(Z<0.959)-P(Z<-0.3196)=0.831-0.375=0.457`

So then the best option is : c. A and C