Question

A recent Pew Center Research survey revealed that 68% of high school students have used tobacco related products. Suppose a statistician randomly selected 20 high school students. Use this information to answer questions 39-41.

1. To find the probability that exactly 15 used tobacco products, would you use the Binomial, Geometric, or Poisson distribution to find the probability?
2. Set up the problem to find P (X = 15) using the appropriate distribution. Do NOT solve.
3. What is the expected number of high school students in the sample who have used tobacco products?

For a self check out at the local Walmart, the mean number of customers per 5 minute interval is 1.5 customers. Use this information to answer questions 42 and 43.

1. To find the probability of 2 customers in the next 5 minutes, would you use the Binomial, Geometric, or Poisson distribution to find the probability?
2. What is the variance (numeric value)?  

Assuming the grades on the first homework are nearly normal with N(90, 4.3), what proportion of grades fall between 85 and 90?
Assuming the grades on the final exam are nearly normal with N(90, 4.3), for a grade of 95 or more on the exam, find the z-score and explain what it means.
Assuming the grades on the final exam are nearly normal with N(90, 4.3), what is the minimum grade putting you in the top 15% of the class?
Assuming the grades on the final exam are nearly normal with N(82, 3.86), what proportion of grades fall between 85 and 90?

Answer 1

1) This is a binomial probability distribution

p = 0.68

n = 20

P(X = x) = nCx * p^x * (1 - p)^{n - x}

P(X = 15) = 20C15 * (0.68)^15 * (0.32)^5 = 0.1599

Expected number of students = np = 20 * 0.68 = 13.6

2) This is a poisson distribution.

= 1.5

P(X = x) = e/x!

P(X = 2) = e^(-1.5) * (1.5)^2/2! = 0.2510

Variance = = 1.5

3) P(85 < X < 90)

= P((85 - )/ < (X - )/ < (90 - )/)

= P((85 - 90)/4.3 < Z < (90 - 90)/4.3)

= P(-1.16 < Z < 0)

= P(Z < 0) - P(Z < -1.16)

= 0.5 - 0.1230

= 0.3770

4) z-score = (X - )/

                = (95 - 90)/4.3

                = 1.16

The grade score 95 is 1.16 standard deviation above the mean.

5) P(X > x) = 0.15

or, P((X - )/ > (x - )/) = 0.15

or, P(Z > (x - 90)/4.3) = 0.15

or, P(Z < (x - 90)/4.3) = 0.85

or, (x - 90)/4.3 = 1.04

or, x = 1.04 * 4.3 + 90

or, x = 94.472

6) P(85 < X < 90)

= P((85 - )/ < (X - )/ < (90 - )/)

= P((85 - 82)/3.86 < Z < (90 - 82)/3.86)

= P(0.78 < Z < 2.07)

= P(Z < 2.07) - P(Z < 0.78)

= 0.9808 - 0.7823

= 0.1985