Question

A survey found that 32​% of consumers from a Country A are more likely to buy stock in a company based in Country​ A, or shop at its​ stores, if it is making an effort to publicly talk about how it is becoming more sustainable. Suppose you select a sample of 200 respondents from Country A. Complete parts​ (a) through​ (d) below.

A. What is the probability that in the​ sample, fewer than 32​%are more likely to buy stock in a company based in Country​ A, or shop at its​ stores, if it is making an effort to publicly talk about how it is becoming more​ sustainable?

B. What is the probability that in the​ sample, between 28​% and 36​%are more likely to buy stock in a company based in Country​ A, or shop at its​ stores, if it is making an effort to publicly talk about how it is becoming more​ sustainable?

C. What is the probability that in the​ sample, more than 28​%are more likely to buy stock in a company based in Country​ A, or shop at its​ stores, if it is making an effort to publicly talk about how it is becoming more​ sustainable?

D. If a sample of 800 is​ taken, how does this change your answers to​ (a) through​ (c)?

If a sample of 800 is​ taken, what is the probability that in the sample fewer than 32​% are more likely to buy stock in a company based in Country​ A, or shop at its​ stores, if it is making an effort to publicly talk about how it is becoming more​ sustainable?

​(Round to two decimal places as​ needed.)

If a sample of 800 is​ taken, what is the probability that in the sample between 28​% and 36​%are more likely to buy stock in a company based in Country​ A, or shop at its​ stores, if it is making an effort to publicly talk about how it is becoming more​ sustainable?

​(Round to two decimal places as​ needed.)

If a sample of 800 is​ taken, what is the probability that in the sample more than 28​%are more likely to buy stock in a company based in Country​ A, or shop at its​ stores, if it is making an effort to publicly talk about how it is becoming more​ sustainable?

.

Answer 1

Proportion, p = 0.32

Sample size, n = 200

Mean, p =  0.32

Standard deviation, σp̂ = √(p*(1-p)/n) = √(0.32 * 0.68 / 200) = 0.0330

a)

P(p̂ < 0.32)

= P((p̂ - p)/σp̂ < (0.32 - 0.32)/0.033)

= P(z < 0)

Using excel function:

= NORM.S.DIST(0, 1)

= 0.50

b)

P(0.28 < X < 0.36)

= P((0.28 - 0.32)/0.033 < (p̂ - p)/σp̂ < (0.36 - 0.32)/0.033)

= P( -1.2127 < z < 1.2127)

= P(z < 1.2127) - P(z < -1.2127)

Using excel function:

= NORM.S.DIST(1.2127, 1) - NORM.S.DIST(-1.2127, 1)

= 0.7747

c)

P(p̂ > 0.28)

= P((p̂ - p)/σp̂ > (0.28 - 0.32)/0.033)

= P(z > -1.2127)

= 1 - P(z < -1.2127)

Using excel function:

= 1 - NORM.S.DIST(-1.2127, 1)

= 0.8874

d)

Mean, p =  0.32

Standard deviation, σp̂ = √(p*(1-p)/n) = √(0.32 * 0.68 / 800) = 0.0165

i)

P(p̂ < 0.32)

= P((p̂ - p)/σp̂ < (0.32 - 0.32)/0.0165)

= P(z < 0)

Using excel function:

= NORM.S.DIST(0, 1)

= 0.50

ii)

P(0.28 ≤ X ≤ 0.36)

= P((0.28 - 0.32)/0.0165 ≤ (p̂ - p)/σp̂ ≤ (0.36 - 0.32)/0.0165)

= P( -2.4254 ≤ z ≤ 2.4254)

= P(z < 2.4254) - P(z < -2.4254)

Using excel function:

= NORM.S.DIST(2.4254, 1) - NORM.S.DIST(-2.4254, 1)

= 0.9847

iii)

P(p̂ ≥ 0.28)

= P((p̂ - p)/σp̂ ≥ (0.28 - 0.32)/0.0165)

= P(z ≥ -2.4254)

= 1 - P(z < -2.4254)

Using excel function:

= 1 - NORM.S.DIST(-2.4254, 1)

= 0.9924