Question

1: Suppose that you take a random sample of 300 people and find that 102 of them say they prefer to buy organic food whenever possible, even if it’s more expensive.

What is the sample proportion of people who prefer to buy organic?

What is the standard deviation of the sample proportion?

Calculate a 95% confidence interval for the population proportion.

Do you reject the null hypothesis: p = 40%? Do you reject the null hypothesis p = 30%?

2: Suppose you work for a car insurance company that covers about 200,000 drivers in New Hampshire. The company estimates that about 14% of people in the state will be involved in a car accident over the next year.

What is the expected number of drivers covered by your company who will be involved in a car accident?

What is the variance of the number of people your company covers who will be involved in car accidents?

Suppose, however, that of those 200,000 drivers you cover, 2,500 are aged 16 – 18, a group for which 18% of drivers will be in an accident in a given year. How many accidents involving 16 – 18 year old drivers would you expect to occur?

Find the probability that someone aged 19 or older will be involved in an accident. (Hint: You can easily find both the number of such drivers your company covers, and the expected value, based on parts A and C).

For #s 3 – 6, suppose that student composite SAT scores (on a 1600 scale) have an average of 1080, a standard deviation of 180, and are normally distributed.

What is the probability that a randomly selected student scores between 1100 and 1200?

What is the probability that a randomly selected student scores between 1400 and 1500?

3.     What is the probability that a randomly selected student scores above 1350?

4.     Suppose you take a random sample of 25 students. What is the probability that the average score of the students in this sample falls between 1100 and 1200? (Hint: The expected value of the sample mean is the population mean, which you were given. The difference between this question and #3 is the standard deviation.)

5. What is the probability that a randomly selected student scores above 1350?

6. Suppose you take a random sample of 25 students. What is the probability that the average score of the students in this sample falls between 1100 and 1200? (Hint: The expected value of the sample mean is the population mean, which you were given. The difference between this question and #3 is the standard deviation.)

Answer 1

1. The random sample of size is n=300 and finds that X =102 of them say they prefer to buy organic food whenever possible. The sample proportion of people who prefer to buy organic is calculated as follows: _102 300 = 0.34 Hence, the sample proportion of people who prefer to buy organic is p = 0.34 The standard deviation of the sample proportion is calculated as follows: P(1-2) Vn 0.34(1-0.34) V 300 =0.02735 Hence, the standard deviation of the sample proportion is 0.02735. The 95% confidence interval for the population proportion is calculated as follows: C1=(p=2,6 f34,) =(0.341.96x0.02735)

= (0.34 0.0536) =(0.2864,0.3936) Hence, the 95% confidence interval for the population proportion is (0.2864,0.3936) The value of p=0.40 is not in the confidence interval, then reject the null hypothesis. The value of p=0.30 is in the confidence interval, then it is failed to reject the null hypothesis. Therefore, 40% 30% Decision Reject the null hypothesis Failed to reject the null hypothesis. Let X be a random variable that denotes the people in the state involved in a car accident and it follows a binomial distribution with parameter n=200,000 and p=0.14. The mean and the standard deviation of the binomial distribution is given below: Mean = np Variance = np(1-P)

The expected number of drivers covered by the company who will be involved in a car accident is calculated as follows: E(X)= np = 200,000x0.14 = 28,000 Hence, the expected number of drivers covered by the company who will be involved in a car accident is 28,000. The variance of the number of drivers covered by the company who will be involved in a car accident is calculated as follows: Var(X)= np(1-P) = 200,000 x 0.14x(1-0.14) = 200,000 x 0.14x0.86 = 24,080 Hence, the variance of the number of drivers covered by the company who will be involved in a car accident is 24,080 Suppose that out of 200,000 drivers cover 2500 are aged 16-18, a group for which 18% of drivers will be in an accident. The expected number of accidents involving 16-18-year-old drivers is calculated as follows: E(X)= np = 2500 x 0.18 = 450 Hence, the expected number of accidents involving 16-18-year-old drivers is 450.

The probability that someone aged 19 or older will be involved in an accident is calculated as follows: 28000-450 200,000 - 2500 27,550 197,500 = 0.1395 Hence, the probability that someone aged 19 or older will be involved in an accident is 0.1395 Let X be a random variable that denotes the student composite SAT scores and it follows a normal distribution with mean u=1080 and standard deviation o=180. The probability that a randomly selected student scores between 1100 and 1200 is calculated as follows: P(1100 P(1100 < X <1200) = P11800 X 1200-P 1100-1080X -4 1200-1080 180 = P(0.11<Z<0.67) =P(Z <0.67)- P(Z <0.11) = 0.7486 -0.5438 =0.2048 Hence, the probability that a randomly selected student score between 1100 and 1200 is 0.2048

7 The probability that a randomly selected student scores between 1400 and 1500 is calculated as follows: P(1400<x<1500)=P[? (1400-1080 X-4 1500-1080 180o 180) _ 3207, 420 1804 180) = P(1.77<<<2.33) = P(Z <2.33) - P(Z<1.77) = 0.9901 -0.9616 From Z-table P(Z<2.33) = 0.9901 and P(Z <1.77)=0.9616) = 0.0285 Hence, the probability that a randomly selected student score between 1400 and 1500 is 0.0285 The probability that a randomly selected student scores above 1350 is calculated as follows: P(X >1350)=1-P(X 51350) =1-P(454,135052080) =1-P(Z 51.5) =1-0.9332 (From Z-table P(Z <1.5)= 0.9332) = 0.0668

Hence, the probability that a randomly selected student score above 1350 is 0.0668 4. The sample size is n=25. The probability that the average score of the students in this fall between 1100 to 1200 is calculated as follows: P(1100<Ē<1200)=P 1100-1080 180 - .< 1200-1080 180 25 20x3 <z<2000) =P(0.55 <Z<3.33) =P(Z<3.33) - P(Z <0.55) = 0.9996 -0.7088 (From Z-table P(Z <3.33)=0.9996 and P(Z <0.55) =0.7088) = 0.2908 Hence, the probability that the average score of the students in this fall between 1100 to 1200 is 0.2908