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  • For the following problems, recall that value = mean + (#ofSTDEVs)(standard deviation) . Verify the mean and standard deviation or a calculator or computer.
  • For a sample: x = x ¯ + (#ofSTDEVs)( s )
  • For a population: x = μ + (#ofSTDEVs)( σ )
  • For this example, use x = x ¯ + (#ofSTDEVs)( s ) because the data is from a sample

  1. Verify the mean and standard deviation on your calculator or computer.
  2. Find the value that is one standard deviation above the mean. Find ( x ¯ + 1s).
  3. Find the value that is two standard deviations below the mean. Find ( x ¯ – 2s).
  4. Find the values that are 1.5 standard deviations from (below and above) the mean.
    • Clear lists L1 and L2. Press STAT 4:ClrList. Enter 2nd 1 for L1, the comma (,), and 2nd 2 for L2.
    • Enter data into the list editor. Press STAT 1:EDIT. If necessary, clear the lists by arrowing up into the name. Press CLEAR and arrow down.
    • Put the data values (9, 9.5, 10, 10.5, 11, 11.5) into list L1 and the frequencies (1, 2, 4, 4, 6, 3) into list L2. Use the arrow keys to move around.
    • Press STAT and arrow to CALC. Press 1:1-VarStats and enter L1 (2nd 1), L2 (2nd 2). Do not forget the comma. Press ENTER.
    • x ¯ = 10.525
    • Use Sx because this is sample data (not a population): Sx=0.715891
  1. ( x ¯ + 1s) = 10.53 + (1)(0.72) = 11.25
  2. ( x ¯ – 2 s ) = 10.53 – (2)(0.72) = 9.09
    • ( x ¯ – 1.5 s ) = 10.53 – (1.5)(0.72) = 9.45
    • ( x ¯ + 1.5 s ) = 10.53 + (1.5)(0.72) = 11.61
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On a baseball team, the ages of each of the players are as follows:

21; 21; 22; 23; 24; 24; 25; 25; 28; 29; 29; 31; 32; 33; 33; 34; 35; 36; 36; 36; 36; 38; 38; 38; 40

Use your calculator or computer to find the mean and standard deviation. Then find the value that is two standard deviations above the mean.

μ = 30.68

s = 6.09
( x ¯ + 2 s ) = 30.68 + (2)(6.09) = 42.86.

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Explanation of the standard deviation calculation shown in the table

The deviations show how spread out the data are about the mean. The data value 11.5 is farther from the mean than is the data value 11 which is indicated by the deviations 0.97 and 0.47. A positive deviation occurs when the data value is greater than the mean, whereas a negative deviation occurs when the data value is less than the mean. The deviation is –1.525 for the data value nine. If you add the deviations, the sum is always zero . (For [link] , there are n = 20 deviations.) So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you make them positive numbers, and the sum will also be positive. The variance, then, is the average squared deviation.

The variance is a squared measure and does not have the same units as the data. Taking the square root solves the problem. The standard deviation measures the spread in the same units as the data.

Notice that instead of dividing by n = 20, the calculation divided by n – 1 = 20 – 1 = 19 because the data is a sample. For the sample variance, we divide by the sample size minus one ( n – 1). Why not divide by n ? The answer has to do with the population variance. The sample variance is an estimate of the population variance. Based on the theoretical mathematics that lies behind these calculations, dividing by ( n – 1) gives a better estimate of the population variance.

Questions & Answers

answer for question 3
Awuah Reply
what are the terminologies in statistics?
Abranti Reply
population sample statistic parameter etc..
the mode of the table class 3_7 8_12 13 _17 frequency 10 25 19
Samah Reply
how touching
i didn't know this app had a chat
I think 25 is higher frequency
compare to 10 and 19
by using formula...?
L+((f1-f0)÷2f1-f0-f2)×h by this formula using ans is 9.8
Me too I think the highest frequency is 25
how is useful
how is stem and leaf plot or display differ from double stem and leaf plot or display
Sebili Reply
What is the variances of 568
Helen Reply
what u asking...
what variance would have a single value..?
variance happened only in a group of values..
if we have a group of values...1st we find its average..ie..'mean'..then we calculate each value's difeerence from the mean..then we will square each 'difference value'.then we devide total of sqared value by n or n-1..that is what variance...
compare the relative adrantages and disadvantages of mean mediam and made
What is the variances of 258
Helen Reply
what is the sample size if the degree of freedom is 25?
Arthur Reply
degrees of freedom may differ with respect to distribution...so tell which distribution you have selected...?
my distribution is 27
what is degree of freedom?
how to understand statistics
Edwin Reply
you are working for a bank.The bank manager wants to know the mean waiting time for all customers who visit this bank. she has asked you to estimate this mean by taking a sample . Briefly explain how you will conduct this study. assume the data set on waiting times for 10 customers who visit a bank. Then estimate the population mean. choose your own confidence level.
Halpha Reply
what marriage for 10 years
Ambaye Reply
fit a least square model of y on x ? what is the regression coefficient ? x : 2 3 6 8 9 10 y : 5 6 7 10 8 11
Nayab Reply
how can we find the expectation of any function of X?
I've been using this app for some time now. I'm taking a stats class in college in spring and I still have no idea what's going on. I'm also 55 yrs old. Is there another app for people like me?
yes I am. it's been decades since I've been in school.
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what is probability
zaheer Reply
simply probability means possibility.. definition:Probability is a measure of the likelihood of an event to occur.
fit a least square model of y on x ? what is the regression coefficient ? x : 2 3 6 8 9 10 y : 5 6 7 10 8 11
classification of data by attributes is called
Yamin Reply
qualitative classification
tell me details about measure of Dispersion
Following data provided Class Frequency less than 10 10-20 5 15 10-30 25 12 40 and above Which measure of central tendency would you compute and why?
Nija Reply

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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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