Question

1. Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 162000 dollars. Assume the standard deviation is 31000 dollars. Suppose you take a simple random sample of 70 graduates.

Find the probability that a single randomly selected policy has a mean value between 153848.5 and 170892.5 dollars.
P(153848.5 < X < 170892.5) =  (Enter your answers as numbers accurate to 4 decimal places.)

Find the probability that a random sample of size n=70n=70 has a mean value between 153848.5 and 170892.5 dollars.
P(153848.5 < X̅ < 170892.5) =  (Enter your answers as numbers accurate to 4 decimal places.)

2.Out of 500 people sampled, 305 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids.

Give your answers as decimals, to three places

< p <

3.If n = 600 and ˆpp^ (p-hat) = 0.3, construct a 99% confidence interval.

Give your answers to three decimals

< p <

4.Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 406 drivers and find that 289 claim to always buckle up. Construct a 86% confidence interval for the population proportion that claim to always buckle up.

Round to 4 decimal places. Interval notation ex: [0.1234,0.9876]

5.In a survey, 15 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $42 and standard deviation of $10. Construct a confidence interval at a 80% confidence level.

Give your answers to one decimal place.

Answer 1