Question

according to www.usmint.gov, 54% of the quarters minted in a recent year were produced by the u.s. mint in denver (the rest were produced in philadelphia). suppose we select a random sample of 200 quarters produced that year. let x = the number of quarters in the sample that were minted in denver.
is the probability distribution of x approximately normal? justify your answer.
np =
and n(1 - p) =
therefore, the probability distribution of x
approximately normal.

Explanation:

Step1: Identify n and p

We know that the sample size \( n = 200 \) (number of quarters in the sample) and the proportion of quarters minted in Denver, \( p = 0.54 \) (since 54% were minted in Denver).

Step2: Calculate \( np \)

To check the normal approximation for a binomial distribution, we first calculate \( np \).
\( np=200\times0.54 = 108 \)

Step3: Calculate \( n(1 - p) \)

Next, we calculate \( n(1 - p) \). First, \( 1-p=1 - 0.54=0.46 \)
Then, \( n(1 - p)=200\times0.46 = 92 \)

Step4: Check normal approximation conditions

For a binomial distribution \( X\sim B(n,p) \), the normal approximation \( X\sim N(np,np(1 - p)) \) is appropriate when \( np\geq10 \) and \( n(1 - p)\geq10 \). Here, \( np = 108\geq10 \) and \( n(1 - p)=92\geq10 \), so the normal approximation is appropriate.

Answer:

\( np=\boldsymbol{108} \), \( n(1 - p)=\boldsymbol{92} \). Since \( np = 108\geq10 \) and \( n(1 - p)=92\geq10 \), the probability distribution of \( X \) is approximately normal.