Question

according to the u.s. census bureau, 24% of u.s. residents are under 18 years old. suppose we select a random sample of 500 u.s. residents. let ( x ) = the number of people in the sample who are under 18 years old.
(a) show that the probability distribution of ( x ) is approximately normal.
the probability distribution of ( x ) is approximately normal because ( np = ) (\boxed{}) and ( n(1 - p) = ) (\boxed{}) are both (\boxed{}).
(b) calculate the mean and standard deviation of the appropriate normal distribution.
mean = (\boxed{}) (do not round.)
standard deviation = (\boxed{}) (round to 2 decimal places.)
(c) use this normal distribution to calculate the probability that the number of people under 18 years old in a random sample of size 500 is between 100 and 110.
(round to 4 decimal places. leave your answer in decimal form.)

Explanation:

Step1: Identify the distribution

We have a binomial distribution problem where \( n = 500 \) (sample size) and \( p=0.24 \) (probability of success, i.e., being under 18 years old). For a binomial distribution \( X\sim B(n,p) \), the conditions for normal approximation are \( np\geq5 \) and \( n(1 - p)\geq5 \).

Calculate \( np \): \( np=500\times0.24 = 120 \)

Calculate \( n(1 - p) \): \( n(1 - p)=500\times(1 - 0.24)=500\times0.76 = 380 \)

Both \( np = 120\) and \( n(1 - p)=380 \) are greater than 5, so the normal approximation is valid.

Step2: Calculate the mean of the normal approximation

For a binomial distribution, the mean \( \mu=np \). Substituting \( n = 500 \) and \( p = 0.24 \), we get \( \mu=500\times0.24=120 \)

Step3: Calculate the standard deviation of the normal approximation

The standard deviation \( \sigma=\sqrt{np(1 - p)} \). Substituting \( n = 500 \), \( p = 0.24 \) and \( 1-p=0.76 \), we have:

\( \sigma=\sqrt{500\times0.24\times0.76}=\sqrt{500\times0.1824}=\sqrt{91.2}\approx9.55 \) (rounded to 2 decimal places)

Step4: Calculate the probability that \( X \) is between 100 and 110

First, we standardize the values \( x_1 = 100 \) and \( x_2=110 \) using the formula \( z=\frac{x-\mu}{\sigma} \)

For \( x = 100 \):

\( z_1=\frac{100 - 120}{9.55}=\frac{- 20}{9.55}\approx - 2.09 \)

For \( x = 110 \):

\( z_2=\frac{110 - 120}{9.55}=\frac{-10}{9.55}\approx - 1.05 \)

We need to find \( P(100<X<110)\approx P(-2.09<Z<-1.05) \)

Using the standard normal table, \( P(Z < - 1.05)=0.1469 \) and \( P(Z < - 2.09)=0.0183 \)

So \( P(-2.09<Z<-1.05)=P(Z < - 1.05)-P(Z < - 2.09)=0.1469 - 0.0183 = 0.1286 \)

Answer:

(a) \( np = 120 \), \( n(1 - p)=380 \) (both are \( \geq5 \), so normal approximation is valid)

(b) Mean \( = 120 \), Standard Deviation \( \approx9.55 \)

(c) Probability \( \approx0.1286 \)