Scores are normally distributed with a mean of 86 and a standard deviation of 14. What is the probability that a random student scored below 72?
Hi,Use a Z Score calculator such as this one:Z Score or Standard Value CalculatorRandom Value = 72Mean (u) = 86Standard Deviation = 14You should get this resultUsing a Z Score lookup table:You will see a Z value of 1 = .3413 or 34.13% of the area under the curve.But… to finish the calculation we need to put all the numbers in the right context.The value of .3413 is the area under the curve from “0” (zero, the normalized average of 86) to the value of 100 or 72 (since normal distribution is symmetrical).The number you want is what is the probability of a random student getting a score below 72 (which is 1 standard deviation away from the average, by the way).What we need to do is calculate the percentage (probability) below 72 by subtracting .3413 from .5000 (.5000 is the probability of what is contained in the lower half of the normal distribution and the .3413 is the probability of what is above 72). So .5000 - .3413 = .1587 or the probability that a student scored below 72 is 15.87%.To calculate the probability of a student scoring 72 or better just add the probabilities above the 72 level which would be .3413 + .5000 (upper half of normal distribution which was not affected by our calculations) and we get .8413 or 84.13% probability a student scored above 72.I hope this helps…
The mean mark of 500 students is normally distributed with 55 & standard deviation is 12. Find the no. of students securing marks above 60 & below 45?
ok so here, n=500 X1=45 and X2=60 mean m=55 std deviation s=12for normal distribution, we need to convert X1 and X2 to Z1 and Z2Z1=(X1-m)/s Z2=(X2-m)/son substituting values we get Z1=-0.83 and Z2=1.25Now we see values of areas A1(-0.83) and A2(1.25), corresponding to Z1 and Z2 from cumulative normal distribution table.A1=1-0.7967=0.2023 (Since Z1 is Negative) A2=0.8944Therefore, number of students below 45=0.2023X500=101 (on rounding off)Number of students above 60=(1-0.8944)X500=53 (on rounding)
The mean and standard deviation of a normal distribution are 150 and 12 respectively.?
Isabelle These are all using the "Empirical Rule" ... a) 138 and 162 138 and 162 are both 1 standard deviation away from the mean. So, the Empirical Rule says the percentage is 68% b) 126 and 174 138 and 162 are both exactly 2 standard deviation away from the mean. So, the Empirical Rule says the percentage is 95% c) 126 and 162 Here you have a mixture of 2 standard deviations and 1 standard deviation, so average ... (1/2)(68 + 95) = 81.5% d) 162 and 174 162 is 1 standard deviation to the right of the mean and 174 is 2 standard deviation to the right of the mean. The area to the right 1 standard deviation is 68/2 = 34% The area to the right 2 standard deviation is 95/2 = 47.5%% Answer = 47.5% - 34% = 13.5% Hope that helped
If the sum of squares of deviations of 15 observations from their mean 20 is 240, then what is the value of coefficient of variation (CV)?
CV = (sd/mean)*100Standard deviation sd = sqrt(240/15)=sqrt(16) = 4Mean = 20CV = (4/20)*100 = 0.2*100 = 20Coefficient of variation = 20This implies that the standard deviation is 20 percentage of the mean
If the coefficient of the variation is 40% and the arithmetic mean is 5 then what is its standard deviation?
As we know, coefficient of variation (c.v.) is as ratio of standard deviation to mean.Thus ,we get: [math]c.v. = \frac{\sigma}{mu}[/math]=> Putting values, we get,=> [math]\frac{40}{100} = \frac{\sigma}{5}[/math]=> [math]\sigma = \frac{40×5}{100}[/math]=> [math]\sigma = 2 .[/math]
The weekly wages of 1000 workers are normally distributed with a mean of Rs. 70 and standard deviation of Rs. 5. What is the number of workers whose weekly wages will be more than 80?
It says it is normal distribution which means the value of the wages follow the curve below. Mean is given as 70 INR so it represents the biggest vertical line in the centre.SD is given as 5 INR so that just means that around 68.26% of workers have their wages between -5 to +5 of the mean (65-75 INR) and 95.44% between -10 to +10 of the mean (60-80 INR)You want to know the number of workers getting more than 80 INR ..that is the region after second standard deviation (the portion that is represented from 2 in the graph above) which is 2.15% + .13% = 2.28%Around 22.8 people (approx 23) :Dhope i was clear in my explanation
If for any moderately asymmetric distribution mode =5 and mean =105, then what is the median?
“Asymmetrical distribution is a situation in which the values of variables occur at irregular frequencies and the mean, median and mode occur at different points. An asymmetric distribution exhibits skewness.”The formula for calculating median is3Median= Mode+2meanBy substituting the given values of mean and mode in the above formula,3Median = 5+2 (105)3Median = 5+210Median = 215/3 = 71.6666666667The value of Median is 71.67.
Guys!!!!! need your help with this Statistics problem:?
The maximum load (with a generous safety factor) for the elevator in an office building is 2000 pounds. The relative frequency distribution of the weights of all men and women using the elevator is mound-shaped (slightly skewed to the heavy weights), with mean μ, equal to 150 pounds and standard deviation equal to 35 pounds. What is the largest number of people you can allow on the elevator if you want their total weight to exceed the maximum weight with a small probability (say, .01)? (HINT: If x1, x2, . . . , xn are independent observations made on a random variable x, and if x has mean μ, and variance 2, then the mean and variance of Pn i=1 are nμ, and n2, respectively. Assume also that the distribution of weights of all men and women using the elevator is normal.)