Download Solutions to Normal Approximation to the Binomial worksheet

MAT 470, Worksheet 6
Name:_____________________
1. Consider a binomial experiment consisting of 100 trials.
The probability of success on each problem is 0.15.
A portion of the corresponding histogram is shown.
Let X be the number of successes.
a). What is the mean and standard deviation
for this distribution?
= 15,  = 12.75  3.5707
b). Determine the following binomial probabilities.
i). P( X = 10 successes ) = 0.04435
ii). P( X = 15 successes ) = 0.111091
c). Use the normal distribution to approximate the following binomial probabilities
i). P( X < 20 successes) = P( Z < 1.54 ) = .9382
compare with
binomcdf(100,0.15,20) = 0.93368
ii). P( X > 18 successes ) = P (Z > 0.98 ) = 0.1635
compare with
1 - binomcdf(100,0.15,18) = 0.16283
iii). P( 8 < X < 22 successes) = P( - 2.1 < Z < 2.1 ) = 0.9642
2. Let X be a normally distributed random variable with mean 60.
Given that P( X < 70) = 0.94, determine the following probabilities:
a). P( X > 70 ) = 0.06
b). P( 60 < X < 70 ) = 0.44
c). P( X < 60 ) = 0.50
d). P( 50 < X < 70 ) = 0.88
3. For a binomial experiment with 200 trials and probability of success p = 0.6, determine
the following values.
a). the mean and the standard deviation for the distribution: = 120,  = 48
b). Using the normal distribution, approximate the probability P( 115 < X < 125 ).
normalcdf(114.5, 125.5, 120, 48 ) = 0.5727
c). Using the normal distribution, approximate the probability P( 110 < X < 130 ).
normalcdf(109.5, 130.5, 120, 48 ) = 0.8704
d). Using the normal distribution, approximate the probability P( X < 130 ).
0.5 + normalcdf(120, 130.5, 120, 48 ) = 0.5 + .435 = 0.935
Note binomcdf(200,0.6,130) = 0.9360974 is the actual cumulative probability.