This is my sample: 29,083 ; 25,314, 31,920 ; 30,750 ; 32,302
I need to find the mean and standard deviation of this sample?!
How can i assure the life spans are normally distributed?
for the mean, add them up and divide by 5
for the sd:
take the difference between each of the data values and the mean and square that result
repeat for all 5 of them
add them up, divide by either 5 or 4 depending on the definition of sd you are using, take the square root.
http://www.wolframalpha.com/input/?i=mean+29083%2C+25314%2C+31920%2C+30750%2C+32302
http://www.wolframalpha.com/input/?i=standard+deviation+29083%2C+25314%2C+31920%2C+30750%2C+32302
thank you but how can I assume that the life spans are normally distributed?
To find the mean and standard deviation of a sample, follow these steps:
1. Add up all the values in the sample: 29,083 + 25,314 + 31,920 + 30,750 + 32,302 = 149,369.
2. Divide the sum by the number of values in the sample: 149,369 / 5 = 29,873.8. This is the mean of the sample.
3. Subtract the mean from each value in the sample, square the result, and sum up these squared differences.
- (29,083 - 29,873.8)^2 = 639,702.44
- (25,314 - 29,873.8)^2 = 21,340,278.24
- (31,920 - 29,873.8)^2 = 4,184,570.24
- (30,750 - 29,873.8)^2 = 773,501.44
- (32,302 - 29,873.8)^2 = 5,923,305.04
4. Divide the sum of squared differences by the number of values in the sample, and then take the square root of this result to find the standard deviation:
-[( 639,702.44 + 21,340,278.24 + 4,184,570.24 + 773,501.44 + 5,923,305.04) / 5]^(1/2) = 5,082.21 (rounded to two decimal places)
To determine if the life spans are normally distributed, you can perform the following checks:
1. Visual inspection: Plot a histogram or a frequency polygon of the life spans and observe if they resemble the typical bell-shaped curve of a normal distribution.
2. Use a normality test: There are statistical tests, such as the Shapiro-Wilk test or the Kolmogorov-Smirnov test, that can formally test the assumption of normality based on the sample data. These tests provide a p-value, and if the p-value is greater than a predetermined significance level (e.g., 0.05), then you can assume that the life spans are normally distributed.
Keep in mind that these checks only provide an indication, and normality cannot be conclusively proven based on a sample. Collecting a larger sample or consulting with a statistician may be necessary for a more accurate assessment.