Question

18#Suppose that 16% of the population of the U.S. is left-handed. If a random sample of 190 people from the U.S. is chosen, approximate the probability that at least 29 are left-handed. Use the normal approximation to the binomial with a correction for continuity.Round your answer to at least three decimal places. Do not round any intermediate steps.

Answer 1

Answer:

0.707

Explanation:

Given that:

This is a binomial distribution having:

16% of the population is left-handed

Sample = 190 people

The probability that at least 29 are left-handed is given as shown below:

The probability of having exactly x successes on repeated trials n, with probability p can be approximated using the standard deviation and expected value. We have:

$\begin{gathered} p=16\text{\%}=0.16 \\ n=190 \\ \text{For binomial distribution, we have:} \\ E(X)=np \\ E(X)=\mu \\ \mu=190\times0.16=30.4 \\ \mu=30.4 \\ \text{The standard deviation for a binomial distribution is:} \\ \sigma=\sqrt{np(1-p)} \\ \sigma=\sqrt{190\times0.16\times(1-0.16)} \\ \sigma=\sqrt{190\times0.16\times0.84} \\ \sigma=\sqrt{25.536}=5.0533157431532021775129883362213 \\ \sigma=5.0533157431532021775129883362213 \end{gathered}$

We will proceed to obtain the z-score as shown below:

$\begin{gathered} \text{ The probability that at least 29 are left-handed} \\ \text{Using continuity correction, we have:} \\ P(X\ge29-0.5)=P(X\ge28.5);\text{ this is the pvalue of Z when X = 28.5} \\ Z=\frac{X-\mu}{\sigma} \\ Z=\frac{28.5-30.4}{5.0533157431532021775129883362213} \\ Z=\frac{-1.9}{5.0533157431532021775129883362213} \\ Z=-0.37599075469889897154114496549266 \\ p-value≈0.706924=0.707 \\ p-value=0.707 \end{gathered}$

Therefore, there is a probability that 0.707