Number of results: 10,402
A fitness company surveyed the weights of its 90 costumers. Assume that the weights are normally distrubted with a mean of 162 lbs and a standard deviation of 3.5 lbs. If you select one of its costumers randomly, what is the probability that the costumer's weight is a) more ...
Friday, November 14, 2008 at 10:55pm by devon
a) 50% b) 97.7% c) 99.9% d) 81.9% These answers were obtained using the normal disribution calculation tool at http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html I did not make use of the fact that the mean and standard deviation were obtained from 90 samples...
Friday, November 14, 2008 at 10:55pm by drwls
Instant dinner comes in packages with weights that are normally distributed, with a standard deviation of 0.3 oz. If 2.3% of the dinners weigh more than 13.5 oz, what is the mean weight?
Saturday, November 15, 2008 at 2:11pm by missy
As Damon suggested, these are done using Normal distribution tables. Personally I don't see the difference in using tables or computer programs that do the same thing. My personal choice is http://davidmlane.com/hyperstat/z_table.html If I recall correctly, there is a ...
Saturday, November 15, 2008 at 2:13pm by Reiny
If the variance is 0.0004, the standard deviation is the square root of that, or 0.02. Using the computational tool at http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html , the probability is 9.1%
Saturday, November 15, 2008 at 2:05pm by drwls
Instant dinner comes in packages with weights that are normally distributed with a standard deviation of 0.3 oz. If 2.3 percent of the dinners weigh more than 13.5 oz, what is the mean weight?
Saturday, November 15, 2008 at 5:49pm by missy
For a class of 43 students fora certain exam, the mean grade is 67 and the standard deviation is 7. a) How many students should recieve a C? (a c is between 74-79) b) How many students should recieve a B? (a B is between a 80-89) c) what score is necessary to get an A?
Saturday, November 15, 2008 at 8:26pm by joe-c
2.3% = .023 so that is the probability that it will be over 13.5 oz, or .977 is the prob that it will be below 13.5 normal distribution charts show the probability below a given z-score. z-score = (real score - mean)/standard deviation I used this chart http://www.math.unb.ca...
Saturday, November 15, 2008 at 5:49pm by Reiny
the standard deviation is the square root of the variance.
Saturday, November 15, 2008 at 2:13pm by Damon
A class of 43 students took an exam. They had a mean average of 67 and a standard deviation of 7. a) how many students got a c b) how many students got a b c) what is a scoree needed to get an a is a) 4 b) 1 c) 90
Sunday, November 16, 2008 at 1:12pm by joe-c
Find probability that a randomly selected TV will have replacement time less than 6 yrs. mean is 8.2 yrs standard deviation is 1.1 then provide warranty of 1% will be replaced what is the time length of the warranty
Monday, November 17, 2008 at 11:03am by Valerie
How large a sample should be taken if the population mean is to be estimated with 99% confidence to within $72? The population has a standard deviation of $800 Please Help, im stuck
Monday, November 17, 2008 at 1:47pm by ks
A study was conducted to estimate the mean amount spent on Christmas gifts for a typical family having two children. A sample of size 150 was taken, and the mean amount spent was $225. Assuming a standard deviation equal to $50, find a 95% confidence interval for , the mean ...
Tuesday, November 18, 2008 at 3:03pm by Kennon
How large a sample should be taken if the population mean is to be estimated with 99% confidence to within $72? The population has a standard deviation of $800.
Tuesday, November 18, 2008 at 3:04pm by Kennon
mu = bar x +/- 2.58(sigma/sqrt N) mu is the population mean. bar x is the average. 2.59 gives the 99% confidence interval. sigma is standard deviation. N is the number of sample. Solve for N.
Tuesday, November 18, 2008 at 3:04pm by DrBob222
A study found that 8-to 12-year olds spend an average of $18.50 per trip to mall. If a sample of 49 children was used, find the 99% confidence inteval of the mean. Assume the standard deviation is $1.50.
Thursday, November 20, 2008 at 11:49pm by Anonymous
The standard deviation from the mean for the average of 49 students will be 1.50/sqrt49 = $0.214. The 99% confidence inverval is $17.95 to $19.05 I used the normal distribtion tool at http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html
Thursday, November 20, 2008 at 11:49pm by drwls
Stats: need help please
Add the mean scores (giving 170). The standard deviation of the total score is the square root of the sum of the squares (5.2)
Sunday, November 23, 2008 at 6:30pm by drwls
1. A certain population follows a normal distribution, with mean m and standard deviation s = 2.5. You collect data and test the hypotheses H0: m = 1, Ha: m 1. You obtain a P-value of 0.022. Which of the following is true? Why? a. A 95% confidence interval for m will include ...
Saturday, November 29, 2008 at 8:32am by Latoya
2. The time (in number of days) until maturity of a certain variety of hot pepper is normally distributed, with mean m and standard deviation s = 2.4. This variety is advertised as taking 70 days to mature. I wish to test the hypotheses H0: m = 70, Ha: m > 70, so I select a...
Sunday, November 30, 2008 at 6:19pm by Latoya