MATH 222 - STATISTICS

On weekday afternoons, the Broadway Avenue Bus (CTA Route 36) runs every 10 minutes, and the waiting times for buses are uniformly distributed. Use this information to fill in the blanks and answer the questions: a) What is the mean waiting time for a Broadway Avenue bus? μ = minutes b) Find the probability that a randomly selected passenger waits less than 6 minutes. (Round answer to three decimal places) probability = c) Find the probability that a randomly selected passenger waits more than 8 minutes.(Round answer to three decimal places) probability = d) Find the third quartile for waiting times. X = minutes

Let X = waiting time of a randomly selected passenger (in minutes).Then we are given that X ~ U(0, 10) a) μ = (0 + 10)/2 = 5 minutes b) The density function is f(x) = 1/10, and we calculate probability as the area under the curve. => P(X < 6) = base * height = 6 * (1/10) = 6/10 or 0.600 c) Again, the probability is calculated as the area under the curve f(x) = 1/10: => P(X > 8) = base * height = 2 * (1/10) = 2/10 or 0.200 d) Here we want to find a value k so that P(X < k) = .75. I.e. P(X < k) = base * height = .75 So k * 1/10 = .75 Multiply both sides by 10 to get k = 7.5

Suppose the value of a stock varies each day from $12 to $27 with a uniform distribution. Find μ, the mean stock price.

19.50 Given a uniform distribution X~U(a, b), the mean is m = (a + b)/2

The following table shows the distribution of family size in a certain U.S. city Family Size Probability 2 0.405 3 0.239 4 0.203 5 0.097 6 0.040 7 0.016 What is the expected number of family members for a family in this city?

3.2 E(X) = S x*P(x) = 2(0.405) + 3*(0.239) + 4(0.203) + 5*(0.097) + 6*(0.040) + 7*(0.016) = 3.176, which rounds to 3.2.

The table below shows approximate probabilities for scoring 0, 1, 2, 3, or 4 runs in a single inning of Major League Baseball. (The probability of scoring more than 4 runs in one inning is extremely small and is not considered here) x P(x) 0 0.83 1 0.09 2 0.04 3 0.03 4 0.01 Use this to fill in the blank and answer the following questions. Give answers to two decimal places accuracy. a) Calculate the probability fewer than 3 runs are scored. P(X < 3) = b) Calculate the probability that at least 2 runs are scored. P(X > 2) = c) Suppose that a single inning of a MLB game is observed; what is the expected number of runs scored?

a) 0.96 Add the probabilities corresponding to X = 0, 1 and 2 b) 0.08 Add the probabilities corresponding to X = 2, 3 and 4 c) 0.30 In the TI-84, go to Stat >> Edit. Enter the x-values into L1 and the P(x) values into L2. Go to Stat >> CALC and select 1-VarStats. Specify L1 as the x-list and L2 as the Freq list, and press CalculateThe expected value appears as x-bar at the top of the screen

x P(x) 0 0.176 1 0.294 2 0.353 3 0.118 4 0.059 Find the mean and standard deviation of the distribution.

mean: 1.59; standard deviation: 1.09 Enter the x-values into L1 Enter the P(x) values into L2. Then go to STAT >> CALC >> 1-Var Stats. Type L1,L2, and hit Enter. The mean variance will be x-bar; the standard deviation is σx.

The probability that a house in an urban area will be burglarized is 3%. If 25 houses are randomly selected, what is the probability that none of the houses will be burglarized?

0.467 This is a binomial distribution problem, where n = 25, p = .03, and X = # houses that were burglarized. The probability that no houses in the sample are burglarized is: P(X = 0) = binompdf(25, .03, 0) = 0.4670.

A recent survey found that 70% of all adults over 50 wear glasses for driving. In a random sample of 10 adults over 50, what is the probability that at least six wear glasses?

0.850 This is a binomial distribution problem, where n = 10, p = .70, and X = # who wear glasses. The probability that the student passes is: P(X > 6) = 1 - P(X < 5) = 1 - binomcdf(10, .70, 5)= 1 - .1503= 0.8497.

Suppose that the value of stock varies each day from $14 to $25 with a uniform distribution. Find the 20th percentile.

16.2 The density function is f(x) = 1/(b − a) = 1/11 We want k so that P(x < k) = 0.20That is, we want the area to the left of k to be .20 => Area = Base*height = 0.20 => (k − 14)(1/11) = .20 => k − 14 = 2.2 Thus, k = 16.2

In 2020, it was found that the percentage of U.S. households earning over $100,000 per year was estimated to be 31%. Suppose that a random sample of 80 households are selected. Use this information to fill in the blanks and answer the questions; for parts a, b and c, round probabilities accurately to four places. a) Find the probability that exactly 20 of the selected households have annual income above $100,000. Probability = b) Find the probability that at most 20 of the selected households have annual income above $100,000. Probability = c) Find the probability that at least 15 of the selected households have annual income above $100,000. Probability = d) Suppose now that a random sample of 500 of households isselected. How many of them would we expect to have annualincomes above $100,000? Expected number =

This problem will involve a Binomial Distribution: - We can view the selection of 80 households as selecting a single household 80 times in succession. Thus, there are n = 80 identical trials. - For each trial, there is exactly two possible outcomes - the household has income above $100k or it does not - The trials are independent of one another - the probability that household has income above $100k does not depend on whether or not previously selected households had income above this level - For each trial, the probability of a "success" (i.e. the probability that a household has income above $100k) is the same: p = 0.31. - Finally, the random variable here is X = number of households in the sample having income above $100k Thus we have a Binomial Distribution with n = 80 and p = .31 a) P(X = 20) = binompdf(80, .31, 20) = 0.0509 b) P(X < 20) = binomcdf(80, .03, 20) = 0.1489 c) If we are interested in the outcomes with 15 or more successes then the highest number of successes we do not want would be 14. By the complement rule, we have: P(X > 15) = 1 - P(X < 14) = 1 - binomcdf(80, .31, 14) = 0.9953 d) The expected value is E(X) = n . p = .31(500) = 155