Wednesday

April 23, 2014

April 23, 2014

**Recent Homework Questions About Statistics**

Post a New Question | Current Questions

**Statistics**

I'll be glad to check your answer. If you get started and get stuck, please post it and tell us what confuses you.
*Saturday, October 19, 2013 at 5:07pm*

**Statistics**

The health information managemnet supervisor agrees to pay a new graduate $13.50 per hour. This is a full-time coding position at 2,080 hours per year. The cost for a full-time employee's fringe benefits is 24 percent of the employee's salary. How much must the ...
*Saturday, October 19, 2013 at 5:03pm*

**Statistics**

13.85 * 1.035
*Saturday, October 19, 2013 at 4:57pm*

**Statistics**

What was the original number of discharges coded each day? Take 15% of that and multiply by 10.
*Saturday, October 19, 2013 at 4:50pm*

**Statistics**

The health information management department at Community Hospital will experience a 15 percent increase in the number of discharges coded per day as the result of opening a cardiac clinic in the facility. The 15 percent increase is projected to be additional rrecords per day...
*Saturday, October 19, 2013 at 4:44pm*

**statistics**

448/450 = 0.995556 = 99.6%
*Saturday, October 19, 2013 at 4:39pm*

**Statistics**

60/20 = 3 per hour for each coder (3 * 7.5) = 22.5 per coder per day 500/5 = 100 discharges a day 100 / 22.5 = 4.44 FTE needed
*Saturday, October 19, 2013 at 4:35pm*

**Statistics**

A coding supervisor must determine the number of full-time equivalents (FTEs) needed to code 500 discharges per week. If it takes an average of 20 minutes to code each record and each coder works 7.5 productive hours per day, how many FETs will the coding supervisor need?
*Saturday, October 19, 2013 at 4:22pm*

**Statistics**

It takes approximately 15 minutes to code one record. How many personnel hours would be needed to code 1,424 discharges for the month?
*Saturday, October 19, 2013 at 4:13pm*

**Statistics**

An employee currently earning $13.85 per hour has been awarded a 3.5 percent merit raise. What will the employee's hourly salary be with this increase?
*Saturday, October 19, 2013 at 3:59pm*

**statistics**

After one month of operation the health information services at community health center determined that there were two reords misfiled out of the 450 active records. What is the filing accuracy rate for this area?
*Saturday, October 19, 2013 at 3:51pm*

**statistics**

Each unit holds 24 feet or 288 inches of shelving. 4,500 * 0.3 = 1,350 inches needed 1350 / 288 = 4.6875 = 5 units Now -- why isn't a new health center starting totally digital? Why is it wasting time, space, and money on out-of-date paper records? Is this why health care ...
*Saturday, October 19, 2013 at 3:46pm*

**statistics**

Community Health center, a new patient clinic in your town serving only underinsured and uninsured patients, has asked you to help its health information management department get started. calculate the number of new shelving units needed for a medical record filing system ...
*Saturday, October 19, 2013 at 3:41pm*

**STAT**

Z = (score-mean)/SD variance = SD^2 Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
*Friday, October 18, 2013 at 8:30pm*

**STAT**

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.04) for your Z score. Z = (score-mean)/SD
*Friday, October 18, 2013 at 8:28pm*

**STAT**

Z = (score-mean)/SD Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z scores.
*Friday, October 18, 2013 at 8:27pm*

**STAT**

Z = (score-mean)/SEm SEm = SD/√n Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
*Friday, October 18, 2013 at 8:22pm*

**statistics**

No, decrease. Mean = ∑x/n
*Friday, October 18, 2013 at 8:19pm*

**statistics**

Use same process as indicated in previous post.
*Friday, October 18, 2013 at 8:00pm*

**statistics**

Use same process as indicated in previous post.
*Friday, October 18, 2013 at 8:00pm*

**statistics**

Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you.
*Friday, October 18, 2013 at 7:58pm*

**statistics**

Which value (1 or 7) is the high value? A, B, C. Z = (score-mean)/SD D. Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to the Z scores.
*Friday, October 18, 2013 at 7:58pm*

**Statistics**

A. Chi square B. Ho: Smoking is not related to SES. H1: Smoking is related to SES. C. Use the Chi-square (X^2) method. X^2 = ∑ (O-E)^2/E, where O = observed frequency and E = expected frequency. ∑ = sum of all the cells. E = (column total * row total)/grand total ...
*Friday, October 18, 2013 at 7:51pm*

**statistics**

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. .6 * (1-.6) = ?
*Friday, October 18, 2013 at 7:27pm*

**statistics**

Either-or probabilities are found by adding the individual probabilities. .23 + .05 = ?
*Friday, October 18, 2013 at 7:26pm*

**Statistics**

See your later post.
*Friday, October 18, 2013 at 7:19pm*

**math**

Z = (score-mean)/SEm SEm = SD/√n Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
*Friday, October 18, 2013 at 7:15pm*

**Statistics**

Z score is your score in arms of standard deviations. Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.6140) related to the Z score. Are you assuming that the scores are ...
*Friday, October 18, 2013 at 7:05pm*

**statistics**

65
*Friday, October 18, 2013 at 6:42pm*

**Statistics **

Serena Williams won the 2010 Wimbledon Ladies Singles Championship. For the seven matches she played in the tournament, her total number of first serves was 379, total number of good first serves was 256, and total number of double faults was 15. a. Find the probability that ...
*Friday, October 18, 2013 at 2:01pm*

**statistics**

.23 + .05 = .28
*Friday, October 18, 2013 at 11:59am*

**statistics**

According to the Current Population Survey of the Census Bureau, 67% of children live with both parents, 23% live with their mother only, and 5% live with their father only. The rest live with neither parent. If we choose a child at random, what is the probability that the ...
*Friday, October 18, 2013 at 11:55am*

**Statistics**

So, how would I figure out the standard deviation of 61.4?
*Thursday, October 17, 2013 at 6:19pm*

**Statistics**

In probability and statistics, mean is used to refer to one measure of the central tendency either of a probability distribution or of the random variable characterized by that distribution. In statistics and probability theory, the standard deviation (represented by the Greek...
*Thursday, October 17, 2013 at 6:09pm*

**Statistics**

What is the mean and standard deviation of 63.7%? What is the mean and standard deviation of 61.4%? Using the 68-95-99.7 rule, describe the sampling distribution of 63.7% Using the 68-95-99.7 rule, describe the sampling distribution of 61.4%? Please help I am so lost...
*Thursday, October 17, 2013 at 6:05pm*

**statistics**

mean = np variance = npq Note: q = 1 - p standard deviation = square root of the variance I'll get you started and let you finish the calculations: mean = 175 * .86 = ? variance = 175 * .86 * .14 = ? Note: * means to multiply I hope this will help.
*Thursday, October 17, 2013 at 5:04pm*

**statistics**

It has been reported that 86% of federal government employees use email. If a sample of 175 federal government employees is selected, find the mean, varience, and standard deviation of the number who use email.
*Wednesday, October 16, 2013 at 9:18pm*

**Statistics**

What is the mean and standard deviation of 63.7%? What is the mean and standard deviation of 61.4%? Using the 68-95-99.7 rule, describe the sampling distribution of 63.7% Using the 68-95-99.7 rule, describe the sampling distribution of 61.4%? Please help I am so lost...
*Wednesday, October 16, 2013 at 7:40pm*

**statistics**

According to the Current Population Survey of the Census Bureau, 67% of children live with both parents, 23% live with their mother only, and 5% live with their father only. The rest live with neither parent. If we choose a child at random, what is the probability that the ...
*Wednesday, October 16, 2013 at 4:19pm*

**statistics**

When Trevor and Brian play ping pong, the probability that Trevor will win is 0.6. If Trevor and Brian play two games of ping pong, find the probability that Trevor wins the first game and Brian wins the second. Assume games are independent. Give your answer as a decimal to ...
*Wednesday, October 16, 2013 at 4:05pm*

**statistics**

z = (125-123)/(9.6/sqrt(144) z = 2/(9.6/12) z = 2.5 z-table 1-.0062 =0.9938
*Monday, October 14, 2013 at 8:37pm*

**statistics**

Assume that blood pressure readings are normally distributed with a mean of 123 and a standard deviation of 9.6. If 144 people are randomly selected, find the probability that their mean blood pressure will be less than 125.
*Monday, October 14, 2013 at 7:51pm*

**math **

Our statistics students, as noted were asked to rate their admiration of Hillary Rodham Clinton on a scale of 1 to 7. They also were asked to rate their admiration of Jennifer Lopez and Venus Williams on a scale of 1 to 7. As noted earlier, the mean rating of Clinton was 4.06...
*Monday, October 14, 2013 at 4:11pm*

**math **

A sample of 148 of our statistics students rated their level of admiration for Hillary Rodham Clinton on a scale of 1 to 7. The mean rating was 4.06, and the standard deviation was 1.70. (For this exercise, treat this sample as the entire population of interest.) a. Use these ...
*Monday, October 14, 2013 at 4:11pm*

**Statistics**

In a survey, 55% of the voters support a particular referendum. If 30 voters are chosen at random, find the standard deviation of the number of voters who support the referendum.
*Monday, October 14, 2013 at 4:09pm*

**Statistics**

Smoking habit Socioeconomic status High Low Smoking 15 40 55 Not smoking 25 10 35 Total 40 50 N=90 a. For these ordinal variables, which hypothesis testing is appropriate to test the relationship between SES and smoking habits variable? b. Write the HO and H1 c. Perform the ...
*Monday, October 14, 2013 at 3:42pm*

**Statistics**

2. A researcher wants to find out how students feel about a proposal to ban smoking on campus. She selected THREE groups of “students who currently smoke”, “students who formerly smoke”, and “students who never smoke”. Also, she selects a sample ...
*Monday, October 14, 2013 at 1:30pm*

**statistics**

https://statistics.laerd.com/statistical-guides/standard-score.php
*Monday, October 14, 2013 at 12:53pm*

**statistics **

Give three reasons why z scores are useful.
*Monday, October 14, 2013 at 12:44pm*

**statistics **

A study of the Consideration of Future Consequences(CFC) scale found a mean score of 3.51, with a standard deviation of 0.61, for the 664 students in the sample(Petrocelli, 2003). For the sake of this exercise, let’s assume that this particular sample comprises the ...
*Monday, October 14, 2013 at 12:44pm*

**statistics**

A sample of 148 of our statistics students rated their level of admiration for Hillary Rodham Clinton on a scale of 1 to 7. The mean rating was 4.06, and the standard deviation was 1.70. (For this exercise, treat this sample as the entire population of interest.) a. Use these ...
*Monday, October 14, 2013 at 12:43pm*

**statistics**

Our statistics students, as noted were asked to rate their admiration of Hillary Rodham Clinton on a scale of 1 to 7. They also were asked to rate their admiration of Jennifer Lopez and Venus Williams on a scale of 1 to 7. As noted earlier, the mean rating of Clinton was 4.06...
*Monday, October 14, 2013 at 12:42pm*

**statistics**

1. In an ANOVA, one group has a much larger sample mean than the other. The analyst decides to remove this group and to conduct the analysis on the remaining groups only. Which of the following statements is correct? Select one: a. This analysis is not correct, as now the ...
*Monday, October 14, 2013 at 7:36am*

**Statistics**

69
*Monday, October 14, 2013 at 12:43am*

**statistics**

A sample has a mean of M 40 If a score of X 55 is removed from the sample then the sample mean will increase
*Sunday, October 13, 2013 at 6:27pm*

**STAT**

Given a sample size of 18, with sample mean 660.3 and sample standard deviation 95.9, we are to perform the following hypothesis testing: Null Hypothesis H0 : µ = 700 Research Hypothesis H1 : µ ≠ 700 •What is the test statistics? •At a 0.05 ...
*Sunday, October 13, 2013 at 9:15am*

**math**

Our statistics students, as noted in Exercise 6.32, were asked to rate their admiration of Hillary Rodham Clinton on a scale of 1 to 7. They also were asked to rate their admiration of Jennifer Lopez and Venus Williams on a scale of 1 to 7. As noted earlier, the mean rating of...
*Saturday, October 12, 2013 at 11:23pm*

**math**

A sample of 148 of our statistics students rated their level of admiration for Hillary Rodham Clinton on a scale of 1 to 7. The mean rating was 4.06, and the standard deviation was 1.70. (For this exercise, treat this sample as the entire population of interest.) a. Use these ...
*Saturday, October 12, 2013 at 11:20pm*

**STAT**

The weights of the 100 students in an introductory statistics class are normally distributed, with a mean of 170 pounds and a standard deviation of 5 pounds. How many students would you expect to have weight between 162 pounds and 178 pounds? What is the probability that a ...
*Saturday, October 12, 2013 at 9:26pm*

**College probability and statistics**

3 letters ---> 26^3 3 numbers, assuming it may start with a zero --- 10^4 number of plates = 26^3 (10^4) = 175760000 (about 176 million, long way to go) ii) so only 24 letters, and 8 numbers -- what do you think? iii) look at the letter part first: assume we put CC at the ...
*Saturday, October 12, 2013 at 9:18pm*

**College probability and statistics**

The current license plates in New York State consist of three letters followed by four digits. ( i ) How many possible distinct license plates can there be? ( ii ) How many possible distinct license plates could there be if the letters O and I and the digits, 0 and 1 were ...
*Saturday, October 12, 2013 at 5:30pm*

**statistics**

5x4x3
*Saturday, October 12, 2013 at 2:59pm*

**STAT**

The weights of the 100 students in an introductory statistics class are normally distributed, with a mean of 170 pounds and a standard deviation of 5 pounds. 1)How many students would you expect to have weight between 162 pounds and 178 pounds? 2)What is the probability that a...
*Saturday, October 12, 2013 at 12:32pm*

**STAT**

The weights of the 100 students in an introductory statistics class are normally distributed, with a mean of 170 pounds and a standard deviation of 5 pounds. How many students would you expect to have weight between 162 pounds and 178 pounds?
*Saturday, October 12, 2013 at 12:31pm*

**STAT**

Given a sample size of 18, with sample mean 660.3 and sample standard deviation 95.9, we are to perform the following hypothesis testing: Null Hypothesis H0 : µ = 700 Research Hypothesis H1 : µ ≠ 700 •What is the test statistics? •At a 0.05 ...
*Saturday, October 12, 2013 at 12:27pm*

**statistics**

p(at least 5 have the same month) p(someone shows w/ at least someone else) P(s)+ P(d) = 100 % 100%- P(d) (12 *11*10*9*8)/12^5 P(d) =95040/248832 = .3819 100% -0.3819 = 0.6181 61.81%
*Friday, October 11, 2013 at 5:03pm*

**statistics**

37
*Friday, October 11, 2013 at 4:27pm*

**statistics**

37
*Friday, October 11, 2013 at 4:26pm*

**statistics**

37
*Friday, October 11, 2013 at 4:26pm*

**statistics**

37
*Friday, October 11, 2013 at 4:26pm*

**statistics**

How many friends must you have to guarantee that at least five of them will have birthdays in the same month?
*Friday, October 11, 2013 at 2:35pm*

**statistics**

Hi Shanice, Numbers in total : 1,2,3,4,5,6,7,8 In the first draw : 2 was selected, therefore the probability of drawing a 2 is 1/8. (since there is only one 2 in the list) In the second draw, 3 was selected. The probability is again 1/8. Therefore the overall probability is: 1...
*Friday, October 11, 2013 at 2:44am*

**college statistics**

7. Mary goes into the supermarket. The probability that she buys (a) regular coffee is .30, (b) decaffeinated coffee is .40, and (c) that she buys both is .30. What is the probability that she buys regular coffee, decaffeinated coffee, or both?
*Thursday, October 10, 2013 at 10:51pm*

**college statistics**

2. A researcher has developed a new questionnaire for measuring dominance and would like to estimate the population parameters for the test scores. The questionnaire is administered to a sample of n = 25 subjects. This sample has an average score of mean = 43 with SS = 2400. ...
*Thursday, October 10, 2013 at 10:51pm*

**statistics**

6. Out of eight numbers, what is the probability of drawing a two on the first draw and a three on the second draw using sampling with replacement?
*Thursday, October 10, 2013 at 10:41pm*

**statistics**

5. Find the weighted mean of the following two samples. Sample 1: n = 5 and the mean = 20 Sample 2: n = 15 and the mean = 10.
*Thursday, October 10, 2013 at 10:40pm*

**statistics**

2. A researcher has developed a new questionnaire for measuring dominance and would like to estimate the population parameters for the test scores. The questionnaire is administered to a sample of n = 25 subjects. This sample has an average score of mean = 43 with SS = 2400. ...
*Thursday, October 10, 2013 at 10:39pm*

**statistics**

You can try a proportional one-sample z-test for this one since this problem is using proportions. Here's a few hints to get you started: Null hypothesis: Ho: p = .759 -->meaning: population proportion is equal to .759 (converting the 75.9% to a decimal). Alternative ...
*Thursday, October 10, 2013 at 4:31pm*

**statistics**

Find the operating characteristics of this system. An airport ticket counter has a service rate of 180 per hour, exponentially distributed, with Poisson arrivals at the rate of 120 per hour
*Thursday, October 10, 2013 at 11:44am*

**statistics**

In 2006, 75.9% of first-year students said they used the Internet for research or homework. Administrators found that 168 of an SRS of 200 first-year students used the Internet for research or homework. Is the proportion of first-year students who used the Internet for ...
*Wednesday, October 9, 2013 at 10:29pm*

**statistics**

Z = (score-mean)/SEm SEm = SD/√n Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
*Wednesday, October 9, 2013 at 11:54am*

**statistics**

Assume that a population is normally distributed with a mean of 100 and a standard deviation of 15. Would it be unusual for the mean of a sample of 3 to be 115 or more? Why or why not? how can i write this in a mathematical expression?
*Wednesday, October 9, 2013 at 3:51am*

**Statistics**

SEdiff = √(SEmean1^2 + SEmean2^2) SEm = SD/√n
*Tuesday, October 8, 2013 at 1:51pm*

**Statistics**

Z = (score-mean)/SEm SEm = SD/√n Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the z score.
*Tuesday, October 8, 2013 at 1:45pm*

**statistics**

Z = (mean1 - mean2)/standard error (SE) of difference between means SEdiff = √(SEmean1^2 + SEmean2^2) SEm = SD/√n If only one SD is provided, you can use just that to determine SEdiff. Find table in the back of your statistics text labeled something like "...
*Tuesday, October 8, 2013 at 1:31pm*

**Business Statistics**

Using the telephone numbers listed in your local directory as your population, randomly obtain 20 samples of size 3. From each telephone number identified as a source, take the fourth, fifth, and sixth digits. Calculate the mean of the 20 samples Draw a histogram showing the ...
*Tuesday, October 8, 2013 at 12:12pm*

**statistics (?)**

Is the probability the same for all 4? What does the "it" refer to?
*Tuesday, October 8, 2013 at 12:10pm*

**Statistics**

Compute SEdiff Mean1=23 Mean2=18 SD1 =4 SD2 =6 N1 =9 N2 =6
*Tuesday, October 8, 2013 at 11:42am*

**statistics**

To see whether people are keeping their car tires inflated to the correct level of 35 pounds per square inch (psi), a tire company manager selects a sample of 36 tires and checks the pressure. The mean of the sample is 33.5 psi, and the population standard deviation is 3 psi. ...
*Monday, October 7, 2013 at 8:31pm*

**Statistics**

An April 15, 2002 report in Time Magazine stated that the average age for women to marry in the United States is now 25.9 years of age. If the standard deviation is assumed to be 4.3 years, find the probability that a random sample of 32 U.S. women would show a mean age at ...
*Monday, October 7, 2013 at 5:39pm*

**STAT (?)**

What following statements? Z = (mean1 - mean2)/standard error (SE) of difference between means SEdiff = √(SEmean1^2 + SEmean2^2) SEm = SD/√n If only one SD is provided, you can use just that to determine SEdiff. Find table in the back of your statistics text ...
*Monday, October 7, 2013 at 1:55pm*

**Statistics (incomplete)**

You have no measure of variability. Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities in terms of Z scores. Z scores is your score in terms of standard deviations. Z = (score-...
*Monday, October 7, 2013 at 1:46pm*

**Statistics**

How do I find the deviation of each x-value from the expected value of 0.97? The chart looks like this... x P(x) x-u 0 0.24 ? 1 0.55 ? 2 0.21 ?
*Sunday, October 6, 2013 at 8:53pm*

**Statistics**

1a. Find the mean first = sum of scores/number of scores Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance. Standard deviation = square root of variance Range = highest score...
*Sunday, October 6, 2013 at 2:37pm*

**statistics**

suppose x is a uniform random variable with c=40 ana d= 70 find the probability that a randomly selected observation is between 43 and 65
*Sunday, October 6, 2013 at 12:54pm*

**Statistics**

Try the binomial probability table: n = 10 x = 5 p = .20 q = 1 - p = 1 - .20 = .80 You can use the table, or calculate by hand using the following formula: P(x) = (nCx)(p^x)[q^(n-x)] With your data using the formula: P(5) = (10C5)(.20^5)[.80^(10-5)] I'll let you take it ...
*Saturday, October 5, 2013 at 6:05pm*

**Statistics**

A manufacturer ship toasters in cartons of 10. In each carton, they estimate a 20% chance that one of the toasters will need to be sent back for minor repairs. What is the probability, that in one carton there will be 5 toasters that need repair? I have no idea on where to ...
*Saturday, October 5, 2013 at 5:20pm*

**Statistics**

A small software company bids on two contracts. It anticipates a profit of $60,000 if it gets the larger contract and $25,000 if it gets the smaller contract. The company estimates that there is a 32% chance it will get the larger contract and a 62% chance it will get the ...
*Saturday, October 5, 2013 at 5:17pm*

**Statistics**

1. A simple random sample of FICO credit rating scores is listed below: 714 751 664 789 818 779 698 836 753 834 693 802 a. Find range, variance, and standard deviation. b. As of this writing, the mean FICO score was reported to be 678. Based on these results, is a FICO score ...
*Saturday, October 5, 2013 at 3:55pm*

**Statistics**

How can you have a probability greater than 1? If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. "Forty-six percent of students eat breakfast and also floss their teeth."
*Saturday, October 5, 2013 at 1:37pm*

Pages: <<Prev | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | **17** | 18 | 19 | 20 | 21 | 22 | 23 | Next>>

Post a New Question | Current Questions