Question

An online quiz divides up people according to their usage of the computer. In a survey of 4,001 respondents, 8% were classified as "productivity enhancers" who are comfortable with technology and use the Internet for its practical value. Suppose you select a sample of 400 students at your school and the population proportion of productivity enhancers is 0.08.

a. What is the probability that in the sample, less than 10% of the students will be productivity enhancers ?

b. What is the probability that in the sample, between 6% and 10% of the students will be productivity enhancers ?

c. What is the probability that in the sample, more than 5% of the students will be productivity enhancers．

I would really appreciate it if you could provide me the formulas in excel to calculate them.

Thank you in advance.

a. Mean: 0.08 Standard deviation: 0.0135647

Probability for X <=

　X value: 0.1 Z value:1.4744196 P (<= 0.1): 0.9298157

b Probability for a Range

From X value:0.06 To X value:-0.1 Z value for 0.06: -1.4744196 P(X<= 0.1): 0.9298157Z value for 0.1: 1.47422 P(X<=0.06): 0.0702 P(X<=0.1) : 0.9298 P(0.06<=X<=0.1): 0.8596

c Common Data

Mean: 0.08 Standard Deviation: 0.0135647

Probability for X

X value : 0.05 Z value : -2.211629 P(X>0.05): 0.9865

Answer 1

a)p = 0.08 , n = 400

P(p^< 0.10)

p^ follow normal distribution with mean = 0.08 and

sd = sqrt(0.08 *0.92/ 400) =

0.013565

P(p^ < 0.10)

= P(Z < ( 0.1 - 0.08)/0.013565)

= P(Z < 1.474383 )

= 0.9298

b)

P(0.06 < p^< 0.10)

= 0.8597

c)

P(p^> 0.05)

= 0.9865

a)
	0.1	0.929816
	0.06	0.070184
		0.859631
c)	0.05	0.986504

Excel formulas

a)
	0.1	=NORMDIST(B2,0.08,SQRT(0.08*0.92/400),1)
	0.06	=NORMDIST(B3,0.08,SQRT(0.08*0.92/400),1)
		=C2-C3
c)	0.05	=1-NORMDIST(B5,0.08,SQRT(0.08*0.92/400),1)