Question

About 10% of people in the United States are left handed. Suppose we take a sample of 120 U.S. Residents, and let X = the number of residents who are left handed. 1. Calculate the mean and standard deviation of the sampling distribution of X. Interpret the standard deviation. 2. Calculate the probability that 16 or more individuals in the sample are left handed. Show your work. 3. Interpret your answer from #2.

Answer 1

Answer:

1) $E(X) = np = 120*0.1 =12$

$SD(X) = \sqrt{np(1-p)}= \sqrt{120*0.1*(1-0.1)} =3.286$

For this case we can interpret that the expected variation around the mean is about 3.3 people approximated

2) $P(X\geq 16)$

And we can use the normal approximation given by:

$X \sim N(12, \sigma =3.286)$

And we can use the z score formula given by:

$z = \frac{X -\mu}{\sigma}$

And if we find the z score for 16 we got:

$z = \frac{16-12}{3.286}= 1.217$

And we can find this probability using the normal standard distribution or excel and we got:

$P(Z>1.217) = 0.112$

3) For this case we can conclude that the probability to find 16 or more individuals left handed in a sample of 120 is about 0.112 or 11.2%

Step-by-step explanation:

Let X represent the random variable for the number of residents who are left handed

And for this case we can model this variable with this distirbution:

$X\sim Binom(n= 120 , p =0.1)$

Part 1

The mean is given by:

$E(X) = np = 120*0.1 =12$

And the deviation would be:

$SD(X) = \sqrt{np(1-p)}= \sqrt{120*0.1*(1-0.1)} =3.286$

For this case we can interpret that the expected variation around the mean is about 3.3 people approximated

Part 2

We want to calculate this probability:

$P(X\geq 16)$

And we can use the normal approximation given by:

$X \sim N(12, \sigma =3.286)$

And we can use the z score formula given by:

$z = \frac{X -\mu}{\sigma}$

And if we find the z score for 16 we got:

$z = \frac{16-12}{3.286}= 1.217$

And we can find this probability using the normal standard distribution or excel and we got:

$P(Z>1.217) = 0.112$

Part 3

For this case we can conclude that the probability to find 16 or more individuals left handed in a sample of 120 is about 0.112 or 11.2%