Question

In a neighborhood in a large city, the number of kids in a family can be modeled by the following box model: [0,0,1,1,1,1,1,2,2,2]. Use this box model to answer questions 8 to 11.
Calculate the average of this box model.

Part A options:

	A) 1.8
	B) 2.1
	C) 1.1
	D) 0.7
	E) 0.5

If you were to select 50 families from this neighborhood, how many kids do you expect to find?

Part B options:

	A) 50
	B) 25
	C) 68.75
	D) 35
	E) 55

Suppose the standard deviation of the box model is 0.70. What is the standard error of the number of kids from this sample of 50 families?

Part C options:

	A) 7.7782
	B) 4.9497
	C) 35
	D) 5.1913
	E) 3.2404

What is the probability that the total number of kids in the sample is below 53?

Part D options:

	A) -0.4
	B) 0.6554
	C) 0.61
	D) 0.3446
	E) 0.7291

Answer 1

The box model is [0,0,1,1,1,1,1,2,2,2]

i.e,

X	f(X)=Frequency
0	2
1	5
2	3
Total (n)	10

(8) Thus the average of this box model is :

= (0*2+1*5+3*2)/(2+5+3)= 1.1 (i.e, option C)

(9) For a 50 family of neighborhood The expected no. of children would be 50*E(Z), Where E(Z) is the expected value of children in a family, which is equal to sample mean, i.e, 1.1, Thus 50*E(Z)= 55 (i.e, option E)

(10) The SD is given to be 0.7, The sum of the no. of children in the set of 50 family is given as SD* (this is the sd of sums of all 50 families)= * SD = 7.07 * 0.7 = 4.9497(i.e, option B)

(11) We are to find the probability that P(Z<53)= P(Z-55/(SD*sqrt(n)) < 53-55/(SD*sqrt(n))) = (Normalization is aproximated as the no. of families(=50) is quite high)= 0.3446 (i.e, option D)