Parents randomly give trickortreaters 234 or 5 pieces of can

Parents randomly give trick-or-treaters 2,3,4 or 5 pieces of candy. The number of pieces given is a random variable with the probability density function shown above.

Assume there are 36 trick-or-treaters (Treat this as a random sample)

a.What is the distribution for X the sample mean of pieces per trick-or-treater?

b.Find the probability that the sample mean pieces per trick-or-treater is greater than 3.1 but less than 3.9.

c.What would the sample mean need to be in order to be in the top 10% of possible means?

f(x)={1/4 x=2,3,4,5 0 otherwise

Solution

There are 36 trick or treaters.

Mean = 0.25(2+3+4+5) = 3.5

Variance = 0.25(4+9+16+25) = 21.6

Std dev =4.65

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b) P(3..1<x bar<3.9)

For this we convert to z score and try.

P(3.1-3.5/4.65 <z < 3.9-3.5/4.65)

= P(-0.086<Z< 0.086)

= 2(0.0319)

= 0.0638

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c) Let sample mean be y

P(X>y) = 0.10

z score = 1.28

x score = 3.5+1.28(4.67)

=9.33

Sample mean should be atleast 9.33