Data: Male: n=40, x=69.4,s=11.3. Female: n=40,x=76.3,s=12.5 STATISTICS?
455303
To obtain 95% confidence intervals for the population variance we use the result:\n\n[(n-1) S²] ÷ (sigma²) ~ chi-square(n-1)\n\nWhere sigma² is the population variance\nand S² is the sample variance\n\nFor the males we have n = 40, so using the result we have a chi-square(39).\n\nFor a 95% confidence interval we want the values that contain 95% of the distribution between them (the lower limit will have 2½% below it and the upper limit will have 2½% above it).\n\nUnfortunately my tables don't have a chi-square 39 listed - so you will have to look these up in your tables - I will use the chi-square 40 results just to show you the method.  The chi-square 40 results are 24.43 and 59.34.\n\nNow we have these values (called "critical values") - we stick them into our thrilling result:\n\n[39 × 11.3²]/(sigma²) = 24.43\n\nand:\n\n[39 × 11.3²]/(sigma²) = 59.34\n\nRearranging each of these gives us the sigma²:\n\n(sigma²) = [39 × 11.3²]/24.43 = 203.844\n\n(sigma²) = [39 × 11.3²]/59.34 = 83.922\n\nNow you want the confidence interval for the standard deviation (not the variance), so we need to root these:\n\nsigma = 14.3\n\nsigma = 9.16\n\nSo our 95% confidence interval is (9.16, 14.3).\n\nYou can do a very similar thing for the females and then can compare these two limits to see if there is any overlap between them.\n\nHope this helps.
791569
12.5 g/dl minimum female\n13.5 g/dl minimum male
262206
range = mean +- (t/sqrt(N))*st.dev.\n\nWhere t is found in a chart according to the confidence level in question, N is the number of data points, and st.dev. is the standard deviation of the data set.
609465
Both sexes have an equal number. 24. \nChili's rocks!!!
197333
i think \nx = 2664.173
862471
Your going to compare your mean with the population mean and see if your class really better than the average of the population.\nHo: u = xbar\nHa: u < xbar\n\nalpha = 0.10\none tailed test (of course)\ntherefore: z(table) = -1.28 (i'm not sure)\n\nuse z test since you know the population standard deviation.\n\nz = (x - u)/o/sqrt of n\n\nthat's it.\n\ngood luck
803512
Right i would put the height on the x-axis and weight on the y-axis. Plot the points on 2 graphs. one graph for gender and another graph for year. Then using different colours for each graph (e.g Red for girls, Blue for boys) plot the graph. you should be able to tell the trends.
852665
38 to 72 mm in male and 41 to 68 mm in female.
877862
From one Big Daddy to another:\n\n.75(x) + .9(40 - x) = .78(40)\n.75x + 36 -.9x  = 31.2\n-.15x = -4.8\nx = 32\n\nThus, they should use 32L of the 75% mixture, and 8L of the 90%.
127917
A: 2 PPL HAVE SAME GENDER\nB: 2 PPL HAVE SAME BIRTHDAY\n\n\nP(A) = 90/190\nP(B) = 365/365 X 1/365 = 1/365\n\nP(AnB) = P(A) X P(B) = 90/190 X 1/365 = 0.001
725278
the mean is the same as the average. so just add the three numbers together and divide by 3 (because that's how many numbers there are).\nexample: 12, 45, 63\n12 + 45 + 63 = 120\n120 / 3 = 40\nso the mean is 40
189610
Look at the table of values for the normal distribution.  95% of the values are included within +/- 1.9 SD of the mean.  So:  1.9 x 6.2 = 11.78 and the interval is\n\n     26 - 11.78 <= Mean <= 26 + 11.78, or\n           14.22<= Mean >= 37.78
561755
the question is not clear... base on your question, x might be 1.5
813078
P(2 are women) = \nP(1st is woman) * P(2nd is woman) = \n6/10 * 5/9 = 30/90 = 0.333\n\nP(at least 1 is woman) =\n1 - P(both are men) =\n1 - P(1st is man)*P(2nd is man) = \n1 - (4/10)*(3/9) =\n1 - 12/90 =\n78/90 = 0.867\n\nI'm a little rusty on my "given that..." types of problems, and I'd hate to possibly steer you wrong on your remaining two questions.  But let me just say that I *THINK* the P of your 3rd question is 5/9; and I think the P of the 4th question is the same as the P of your 3rd question.\n\n~ ♥ ~
856003
male to female ratio of total US population: 0.97\nmale to female ratio at birth in US: 1.05\n\nNo offense, but rowdyowl's answer simply isn't right. I think he is just using general reasoning without having the data or technical biological information to reach the correct conclusion.
209966
Set up an equation:\n\n72 = .6 * X\n\n72/.6 = X\n\nX = 120
454512
population sample size (n) = 12\nsquare root of sample size = 3.464\nmean of the data = 11.692\nstandard deviation = .777\n\nIt's too hard to type in math notation so, in words, use the formula: \nmean plus or minus critical k-value multiplied by (standard deviation divided by the square-root of n)\n\nSo for your example, it would be:\n11.692 +/- k(.777/3.464) --> take .777 divided by 3.464\n= 11.692 +/- k(.224) so solving for k -->\n11.692 + k(.224) = 13 \nk = 5.84 \n11.692 - k(.224) = 10\nk = 7.6\n\nThis formula is for the confidence interval for the mean of a normal population. It assumes that this was a simple random sample and is just one of the formulas that you could possibly use. You didn't specify if the data was Normally distributed (I'm not going to check and do everything for you). Just to put this answer in perspective, a confidence level of 99.9 % corresponds to a "k-value" of 3.291