Question

1.)

Forty

individuals independently construct a

98

​%

confidence interval for the mean of a certain population. What is the expected number of individuals who will obtain confidence intervals that actually contain the mean of the​population

2.)Forty percent of the employees in a large corporation are registered in the​ corporation’s fitness program. If 3500 employees are selected at random from this​ corporation, what is the probability between 1410 and​ 1445, inclusive, will be registered in the fitness​ program? (Round to the nearest tenth of a​ percent.)

3.)

A

die

is to be tossed

291

times. Let X represent the number of times the

die

lands with

an odd number

showing. What is the

standard deviation

for​ X? (Round to the nearest​ hundredth.)

Answer 1

(1)

The expected number of individuals who will obtain confidence intervals that actually contain the mean of the​population is given by:

40 X 98/100 = 39.2 = 39 (Round to integer)

So

Answer is:

39

(2)

n = 3500

p = 0.40

q =1 -p = 0.60

= np = 3500 X 0.40 = 1400

To find P(1410X1445):

Applying Continuity Correction:

To find P(1409.5<X<1445.5):

Case 1: For X from mid value to 1409.5:

Z = (1409.5 - 1400)/28.9828

= 0.3278

Table of Area Under Standard Normal Curve gives area = 0.1293

Case 2: For X from mid value to 1445.5:

Z = (1445.5 - 1400)/28.9828

= 1.5699

Table of Area Under Standard Normal Curve gives area = 0.4418

So,

P(1410X1445) = 0.4418 - 0.1293 = 0.3125

So,

Answer is:

0.3125

(3)

n = 291

p = 1/2 = 0.5

So,

q = 1 - p = 0.5

So,

Standard Deviation of X is given by:

So,

Answer is:

8.53