SOLUTION: Roll a die 8 times. What is the probability that the number of 1's will be within 2 standard deviations of the expected number of 1's?

This is a binomial experiment.

We calculate the mean:



That's the expected number of 1's.

[Expected number, is really the expected AVERAGE number.  I.e., 

if you rolled the die 8 times each day for a year, and averaged 

up how many 1's you got each day, you would expect to get an 

average of about 1.333... 1's per day.]

We calculate the standard deviation



2 standard deviations is 

2 standard deviations below the expected number is 


 2 standard deviations above the expected number is 


To get the number of 1's to be between those, we must roll

either 0, 1, 2, or 3 1's

We do that with a TI-83 or TI-84 calculator.

press 2ND then VARS

use down arrow to scroll down to binomcdf( 

    [Warning! Don't scroll to binompdf( by mistake!!!]

press ENTER

depending on how new or old your calculator is,

1. If you see the "wizard", make the wizard read

    binomcdf
trials:8
p:1/6
x value:3
Paste

scroll to "Paste" and press ENTER

you will see binomcdf(8,1/6,3)


2. Or if you have an older model, 

you just see binomcdf(, make it read binomcdf(8,1/6,3)

press ENTER

See: .9693435881

That's the probability of rolling 0,1,2,or 3 1's, which is a number of

1's between 2 standard deviations below the expected number and 2

standard deviations above the expected number. 

Edwin