Let X denote the amount of skittles that come in a standard

Let X denote the amount of skittles that come in a standard size bag of skittles at a movie theater. Assume the distribution of X is normal with mu = 18 and theta = 1. Find the the following probalities.

a) using n = 36, find P(X (bar) > 18.4)

b) using n= 36, find P(17.6 < X(bar) < 18.4)

Solution

X follows Normal with mu = 18 and theta = 1

So, using n=36, Xbar will follow Normal with the same mean mu = 18 but variance theta = 1/36

Hence, Y = (X(bar) - 18)/(1/6) will follow a standard normal distribution with mu = 0 and theta = 1

i.e. Y = 6 (X(bar) - 18)

Now for a) P(X(bar)>18.4) equlas P(6(X(bar) - 18)>6(18.4-18))

= P(Y>2.4) = 1 - P(Y<2.4)

From the standard normal CDF this equals

1- 0.9918 = 0.0082

For b) P(17.6<X(bar)<18.4) = P(X(bar)<18.4) - P(X(bar)<17.6)

From part a) P(X(bar)<18.4) = 0.9918

Now P(X(bar)<17.6) = P(6(X(bar)-18)<6(17.6-18))

= P(Y<-2.4)

by the symmetry of standard normal distribution around 0, this equals

P(Y>2.4) which equlas 0.0082 as calculated in part a

So, P(17.6<X(bar)<18.4) = 0.9918 - 0.0082 = 0.9836