Question

A survey found that 25% 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.

1. ​ What is the probability that in the​ sample, fewer than 25% 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?

The probability is __%?

(Round to two decimal places as needed)

2. What is the probability that in the​ sample, between 20​% and 30​% 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?

The probability is __%?

(Round to two decimal places as needed)

3. What is the probability that in the​ sample more than 20​% 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?

The probability is __%?

(Round to two decimal places as needed)

4. If a sample of 800 is taken, how does this change you answers to (a) through (c)

If a sample of 800 is taken, what is the probability that in the​ sample, fewer than 25% 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?

The probability is __%?

(Round to two decimal places as needed)

If a sample of 800 is taken, what is the probability that in the​ sample, between 20​% and 30​% 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?

The probability is __%?

(Round to two decimal places as needed)

If a sample of 800 is taken, what is the probability that in the​ sample more than 20​% 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?

The probability is __%?

(Round to two decimal places as needed)

Answer 1

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.25
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.25*0.75/200)
=0.0306
a.
the probability that in the​ sample, fewer than 25% 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
P(X < 0.25) = (0.25-0.25)/0.0306
= 0/0.0306= 0
= P ( Z <0) From Standard Normal Table
= 0.5
=50%
b.
the probability that in the​ sample, between 20​% and 30​% 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
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.2) = (0.2-0.25)/0.0306
= -0.05/0.0306 = -1.634
= P ( Z <-1.634) From Standard Normal Table
= 0.05113
P(X < 0.3) = (0.3-0.25)/0.0306
= 0.05/0.0306 = 1.634
= P ( Z <1.634) From Standard Normal Table
= 0.94887
P(0.2 < X < 0.3) = 0.94887-0.05113
= 0.8977
= 89.77%
c.
the probability that in the​ sample more than 20​% 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
P(X > 0.2) = (0.2-0.25)/0.0306
= -0.05/0.0306 = -1.634
= P ( Z >-1.634) From Standard Normal Table
= 0.9489
=94.89%

if sample size is 800 taken,
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.25
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.25*0.75/800)
=0.0153
a.
the probability that in the​ sample, fewer than 25% 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
P(X < 0.25) = (0.25-0.25)/0.0153
= 0/0.0153= 0
= P ( Z <0) From Standard Normal Table
= 0.5
=50%
b.
the probability that in the​ sample, between 20​% and 30​% 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
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.2) = (0.2-0.25)/0.0153
= -0.05/0.0153 = -3.268
= P ( Z <-3.268) From Standard Normal Table
= 0.00054
P(X < 0.3) = (0.3-0.25)/0.0153
= 0.05/0.0153 = 3.268
= P ( Z <3.268) From Standard Normal Table
= 0.99946
P(0.2 < X < 0.3) = 0.99946-0.00054 = 0.9989
=99.89%
c.
the probability that in the​ sample more than 20​% 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
P(X > 0.2) = (0.2-0.25)/0.0153
= -0.05/0.0153 = -3.268
= P ( Z >-3.268) From Standard Normal Table
= 0.9995
=99.95%