Question

The PC Magazine shows that the mean price for liquid crystal display (LCD) computer monitors is...

The PC Magazine shows that the mean price for liquid crystal display (LCD) computer monitors is $200 with a standard deviation of $30. If 9 LCD computer monitors are randomly selected, then the distribution of the sample mean price is

Exactly normal with mean $200 and standard deviation $30.

Exactly normal with mean $200 and standard deviation $10.

Approximately normal with mean $200 and standard deviation $10.

Can't be determined without more information

Homework Answers

Answer #1

Solution

Here option (4) is correct.

Given mean price of LCD =200

Standard deviation =30

A random sample of size n= 9 is selected.

Since sample size is very small and distribution of population is unknown , we can’t apply CENTRAL LIMIT THEOREM.Without knowing distribution of parent populatian and n is large then only we can say the distribution of sample mean is normally distributed by CLT. Otherwise the parent population should be normal.

Hence we can’t determine the distribution of sample mean with the given condition.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The PC Magazine shows that the mean price for liquid crystal display (LCD) computer monitors is...
The PC Magazine shows that the mean price for liquid crystal display (LCD) computer monitors is $200 with a standard deviation of $30. If 9 LCD computer monitors are randomly selected, then the distribution of the sample mean price is
A consumer price analyst claims that prices for liquid crystal display (LCD) computer monitors have a...
A consumer price analyst claims that prices for liquid crystal display (LCD) computer monitors have a mean of $ 200 and a standard deviation of $ 50. 1. What is the probability that a randomly selected LCD computer monitor costs less than $ 210? Assume here that the prices are normally distributed. 2. You randomly selected 16 LCD compute monitors. What is the probability that their mean cost is less than $ 210? Assume here that the prices are normally...
A consumer price analyst claims that prices for liquid crystal display (LCD) computer monitors have a...
A consumer price analyst claims that prices for liquid crystal display (LCD) computer monitors have a mean of $200 and a standard deviation of $48. 1. What is the probability that a randomly selected LCD computer monitor costs less than $210? Assume here that the prices are normally distributed 2. You randomly selected 25 LCD compute monitors. What is the probability that their mean cost is less than $210? Assume here that the prices are normally distributed 3. You randomly...
A magazine includes a report on the energy costs per year for​ 32-inch liquid crystal display​...
A magazine includes a report on the energy costs per year for​ 32-inch liquid crystal display​ (LCD) televisions. The article states that 14 randomly selected​ 32-inch LCD televisions have a sample standard deviation of ​$3.29. Assume the sample is taken from a normally distributed population. Construct 99​% confidence intervals for​ (a) the population variance sigmasquared and​ (b) the population standard deviation sigma. Interpret the results.
A magazine includes a report on the energy costs per year for​ 32-inch liquid crystal display​...
A magazine includes a report on the energy costs per year for​ 32-inch liquid crystal display​ (LCD) televisions. The article states that 1414 randomly selected​ 32-inch LCD televisions have a sample standard deviation of ​$3.843.84. Assume the sample is taken from a normally distributed population. Construct 9898​% confidence intervals for​ (a) the population variance sigmaσsquared2 and​ (b) the population standard deviation sigmaσ. Interpret the results. ​(a) The confidence interval for the population variance is ​(nothing​,nothing​). ​(Round to two decimal places...
It has been determined that the mean amount of time that computer science majors spend on...
It has been determined that the mean amount of time that computer science majors spend on homework each week is approximately normally distributed with a mean of 15.2 hours and standard deviation 3.1 hours. What is the probability that a randomly selected computer science major will spend more than 14.5 hours on homework in a given week?
13) Weights of population of men has  the mean of 170 pounds and the standard deviation of...
13) Weights of population of men has  the mean of 170 pounds and the standard deviation of 27 pounds. Suppose 81 men from this population are randomly selected for a certain study The distribution of the sample mean weight is a)exactly normal, mean 170 lb, standard deviation 27lb b) approximately Normal, mean 170 lb, standard deviation 0.3 lb c)approximately Normal, mean 170 lb, standard deviation 3  lb d)approximately Normal, mean equal to the observed value of the sample mean, standard deviation 27...
The owner of a computer repair shop has determined that their daily revenue has mean $7200...
The owner of a computer repair shop has determined that their daily revenue has mean $7200 and standard deviation $1200. The daily revenue is normally distributed. a) What is the probability that a randomly selected day will have a revenue of at most $7000? b) The daily revenue for the next 30 days will be monitored. What is the probability that the mean daily revenue for the next 30 days will exceed $7500?
The retail price of a particular model of a smartwatch has a skewed distribution with mean...
The retail price of a particular model of a smartwatch has a skewed distribution with mean $220 and standard deviation $15. Suppose that you check the retail prices of this smartwatch at randomly selected 33 stores and calculate the sample mean price.(1) What distribution will the sample mean have in this setting? (a) Exact normal distribution (b) Approximate t distribution (c) Approximate normal distribution (d) Standard normal distribution (e) Exact t distribution (2) What is the probability that the sample...
3. The overhead reach distances of adult females are normally distributed with a mean of 200...
3. The overhead reach distances of adult females are normally distributed with a mean of 200 cm and a standard deviation of 8 cm. a. Find the probability that an individual distance is greater than 209.30 cm. b. Find the probability that the mean for 25 randomly selected distances is greater than 197.80 cm.197.80 cm. c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30? 4. Assume that​ women's heights...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT