Question

Suppose that 16% of the population of the U.S. is left-handed. If a random sample of 170 people from the U.S. is chosen, approximate the probability that fewer than 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:

0.6064 = 60.64% probability that fewer than 29 are left-handed.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

16% of the population of the U.S. is left-handed.

This means that $p = 0.16$

Sample of 170 people

This means that $n = 170$

Mean and standard deviation:

$\mu = E(X) = np = 170*0.16 = 27.2$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{170*0.16*0.84} = 4.78$

Probability that fewer than 29 are left-handed.

Using continuity correction, this is P(X < 29 - 0.5) = P(X < 28.5), which is the pvalue of Z when X = 28.5. So

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

$Z = \frac{28.5 - 27.2}{4.78}$

$Z = 0.27$

$Z = 0.27$ has a pvalue of 0.6064

0.6064 = 60.64% probability that fewer than 29 are left-handed.