Asked by Vanessa
Assume a binomial probability distribution has p = .60 and n = 200.
c. What is the probability of 100 to 110 successes (to 4 decimals)?
d. What is the probability of 130 or more successes (to 4 decimals)?
c. What is the probability of 100 to 110 successes (to 4 decimals)?
d. What is the probability of 130 or more successes (to 4 decimals)?
Answers
Answered by
Damon
well, you have enough data points to use normal distribution rather than computing with binomials forever and a day. Otherwise use a program or spreadsheet to compute all the binomial coefs and powers and products
mean = n p = 200 * .6 = 120
sigma^2 = 120 (.4) = 48 so sigma = 6.93
for 100
z = (100 - 120)/6.93 = -20/6.93 = -2.89
for 110
z = (110-120)/6.93) = -1.45
so go to normal distribution table and find from -2.89 to -1.45
about .06
for 130
z =10/6.93 = +1.45
up to 1.45 in the table = .925
so about 1 - .925 = .075
I do not have an accurate table handy so you will have to do it more accurately.
mean = n p = 200 * .6 = 120
sigma^2 = 120 (.4) = 48 so sigma = 6.93
for 100
z = (100 - 120)/6.93 = -20/6.93 = -2.89
for 110
z = (110-120)/6.93) = -1.45
so go to normal distribution table and find from -2.89 to -1.45
about .06
for 130
z =10/6.93 = +1.45
up to 1.45 in the table = .925
so about 1 - .925 = .075
I do not have an accurate table handy so you will have to do it more accurately.
Answered by
Damon
Here is an online normal distribution calculator:
http://davidmlane.com/hyperstat/z_table.html
http://davidmlane.com/hyperstat/z_table.html
Answered by
Damon
by the way that calculator uses x - mean for z instead of (x-mean)/sigma
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