Question

In: Math

A population of young people was studied where the variable weight has an average of 60...

A population of young people was studied where the variable weight has an average of 60 kilograms. The standard deviation is 5 kilograms, and the serura variable presented an average cone of 1.70 meters and its standard deviation 10 centimeters. Calculate the probability of finding young people weighing over 58 kilograms and measuring less than 1.80 meters.

Expert Solution

Here

We have to find P(X > 58)

where z is standard normal variable.

= P(z > -0.4)

= 1 - P(z < -0.4)

= 1 - 0.3446 (From statistical table of z values)

= 0.6554

= P(z < 0.01)

= 0.5040 (From statistical table of z values)

The probability of finding young people weighing over 58 kilograms and measuring less than 1.80 meters

= 0.6554 * 0.5040

= 0.3303

The probability of finding young people weighing over 58 kilograms and measuring less than 1.80 meters is 0.3303

milcah answered 10 months ago

In a group of 12 young professionals studied, it was found that they purchased on average...

In a group of 12 young professionals studied, it was found that they purchased on average $245.804 per month eating meals prepared outside the home, with a standard deviation of $34.4307. What is the 95% confidence interval of the true amount of money young professionals spend monthly on meals prepared outside the home? Question 1 options: 1) ( 224.146 , 267.462 ) 2) ( 223.928 , 267.68 ) 3) ( -223.928 , 267.68 ) 4) ( 235.865 , 255.743 )...

Imagine a population in the midst of an epidemic where 60% of the population is considered...

Imagine a population in the midst of an epidemic where 60% of the population is considered susceptible, 10% are infected, and 30% are recovered. There is only one test for the disease. For susceptible patients the test is accurate (i.e. returns a negative result) 95% of the time. For infected patients the test is correct (i.e. returns a positive result) 99% of the time. The test is correct (i.e. returns a negative results) 65% of the time for recovered individuals....

Assume the average weight of a player in the NFL is normally distributed with a population...

Assume the average weight of a player in the NFL is normally distributed with a population mean of 245 pounds with a population standard deviation of 46 pounds. Suppose we take a sample of 50 NFL players. a. What is the probability that a randomly selected player will weigh over 300 pounds? b. What is the probability that a randomly selected player will weigh under 180 pounds? c. What is the probability that a randomly selected player will weigh between...

Suppose height and weight of people in a certain population are jointly normal with the mean...

Suppose height and weight of people in a certain population are jointly normal with the mean height of 180cm and mean weigh of 80kg. The corresponding standard deviations are 5cm and 6kg, respectively, and the correlation is equal to 0.6. a) What is the percentage of people whose weight exceeds 95kg among those 185cm tall? b) What is the percentage of people whose weight exceeds 95kg among those whose height is 175cm?

A firm has a fixed cost of $100 and average variable cost of $5xq, where q...

A firm has a fixed cost of $100 and average variable cost of $5xq, where q is the number of units produced. a. Construct a table showing total cost for q from 0 to 10. b. Graph the firms marginal cost and average total cost curves. c. How does marginal cost change with q? What does this suggest about the firm’s production process?

In a population of 158 people: 85 men and 73 women, it was found that 60...

In a population of 158 people: 85 men and 73 women, it was found that 60 men and 46 women had health insurance, the rest no. 40 of the men with health insurance and 9 of the men without insurance, visited the MD office more than 5 times last year. In the other hand, 35 of the women with insurance and 8 of the women without insurance, visited the MD office more than 5 times last year. Giving this information,...

A sample of 64 cows is used to estimate the average weight of a herd’s population....

A sample of 64 cows is used to estimate the average weight of a herd’s population. If the population standard deviation is 96 pounds, what is the margin of error of the population average weight estimate for a 90% confidence interval?

1.A report in a research journal states that the average weight loss of people on a...

1.A report in a research journal states that the average weight loss of people on a certain drug is 23 lbs with a margin of error of ±2 lbs with confidence level C = 95%. (a) According to this information, the mean weight loss of people on this drug, μ, could be as low as .........lbs. (b) If the study is repeated, how large should the sample size be so that the margin of error would be less than 1...

A medical statistician wants to estimate the average weight loss of people who are on a...

A medical statistician wants to estimate the average weight loss of people who are on a new diet plan. In a study, he guesses that the standard deviation of the population of weight losses is about 20 pounds. How large a sample should he take to estimate the mean weight loss to within 8 pounds from the sample average weight, with 95% confidence? Is your estimation always correct? Please briefly explain your reasoning. If you start with a bigger sample,...

Sophie has a business where she sells DVDs for $15 each. The average variable cost of...

Sophie has a business where she sells DVDs for $15 each. The average variable cost of production is $10 per unit. The average total cost of production is $16. What should she do in the short run? a. She should continue operating since the business is earning an economic profit. b. She should shut down because the sales are NOT covering the average total cost of production. c. She should shut down since the business is earning zero economic profit....