Question

The median age of residents of the United States is 31 years. If a survey of 100 randomly selected U.S. residents is to be taken, use the normal approximation to the binomial distribution to approximate the probability that at least 57 will be under 31 years of age.

Answer 1

To approximate the probability that at least 57 out of 100 randomly selected U.S. residents will be under 31 years of age, we can use the normal approximation to the binomial distribution. The probability is approximately 8.08%.

Step-by-step explanation:

To approximate the probability that at least 57 out of 100 randomly selected U.S. residents will be under 31 years of age, we can use the normal approximation to the binomial distribution.

First, we need to find the mean and standard deviation for the binomial distribution. The mean (μ) is given by n * p, where n is the number of trials (100 in this case) and p is the probability of success (being under 31 years of age). The standard deviation (σ) is given by sqrt(n * p * (1-p)).

Using the formula above, the mean is 100 * 0.5 = 50, and the standard deviation is sqrt(100 * 0.5 * (1-0.5)) = 5. Therefore, we have a normal distribution with a mean of 50 and a standard deviation of 5.

To find the probability that at least 57 out of 100 residents will be under 31 years of age, we can use the cumulative distribution function (CDF) of the normal distribution. We want to find the probability that the random variable is greater than or equal to 57, which is equivalent to finding the probability that the random variable is less than or equal to 43 (since the distribution is symmetric).

Using a standard normal distribution table or a calculator, we can find the Z-score corresponding to 43 by subtracting the mean from 43 and dividing by the standard deviation. The Z-score is (43-50)/5 = -1.4. The probability of getting a Z-score less than or equal to -1.4 is approximately 0.0808. Therefore, the probability that at least 57 out of 100 residents will be under 31 years of age is approximately 0.0808 or 8.08%.

Answer 2

P (x≥ 57) = 6.7789 e^-8

Step-by-step explanation:

Here n= 100

p = 31/100 = 0.31

We formulate the null hypothesis that H0: p= 0.31 against the claim Ha: p≠0.31

The significance level is chosen to be ∝= 0.05

The test statistic x to be used is X, the number U.S. residents is to be taken which is at least 57

The binomial calculator gives the

P (x≥ 57) = 6.7789 e^-8

IF ∝= 0.05 then ∝/2 = 0.025

We observe that P (x≥ 57) is less than 0.025

Hence we reject H0 and conclude that p ≠0.31

This is true because for normal distribution the median = mean which is usually the 50 % of the data.