Question

The number of meals per day, X, Sarah consumes, has the following distribution.

x	2	3	4	5
p(x)	0.25	0.35	0.25	0.15

Assume that the number of meals Sarah consumes is independent is from day to day.

A. If two days are selected at random, what is the sampling distribution of the mean?

B. Find the expected value and variance for both X and X?.

C. If 40 days are randomly selected, what is the probability that the sample mean is between 3 and 3.5?

Answer 1

a)

below is sampling distribution of mean:

P(Xbar=2)=P(x1=2)*P(X2=2)=0.25*0.25=0.0625

P(Xbar=2.5)=P(X1=2)*P(X2=3)+P(X1=3)*P(X2=2)=0.25*0.35+0.35*0.25=0.175

P(Xbar=3)=0.2475

P(Xbar=3.5)=0.25

P(Xbar=4)=0.1675

P(Xbar=4.5)=0.075

P(Xbar=5)=0.0225

b)

for X:

mean =3.3

variance =1.010

for sample mean Xbar:

mean =3.3

variance =0.505

c)

for 40 days ; expected mean =3.3

and std deviation =sqrt(variance/n)=sqrt(1.01/40)=0.159

therefore from normal approximation :

probability that the sample mean is between 3 and 3.5 =P(3<Xbar<3.5)

=P((3-3.3)/0.159<Z<(3.5-3.3)/0.159)=P(-1.89<Z<1.26)=0.8962-0.0294=0.8668