Question

A survey found that 22% 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 100 respondents from Country A. Complete parts (a) through (d) below. a. What is the probability that in the sample, fewer than 22% 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 50 %. (Round to two decimal places as needed.) b. What is the probability that in the sample, between 19% and 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 53.11 %. (Round to two decimal places as needed.) c. What is the probability that in the sample, more than 19% 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

Solution

Given that,

p = 0.22

1 - p = 1 - 0.22 = 0.78

n = 100

$\mu$$\hat p$ = p = 0.22

$\sigma$$\hat p$  $\sqrt$[p ( 1 - p ) / n] = $\sqrt$ [(0.22 * 0.78) / 100 ] = 0.0414

b) P( 0.19 < $\hat p$ < 0.25 )

= P[(0.19 - 0.22) / 0.0414 < ( $\hat p$ - $\mu$$\hat p$ ) / $\sigma$$\hat p$ < (0.25 - 0.22) / 0.0414 ]

= P(-0.72 < z < 0.72)

= P(z < 0.72) - P(z < -0.72)

Using z table,   

= 0.7642 - 0.2358

= 0.5284

The percentage is = 52.84%

c) P( $\hat p$ > 0.19) = 1 - P( $\hat p$ < 0.19 )

= 1 - P(( $\hat p$ - $\mu$$\hat p$ ) / $\sigma$$\hat p$ < (0.19 - 0.22) /0.0414 )

= 1 - P(z < -0.72)

Using z table

= 1 - 0.2358

= 0.7642

The percentage is = 76.42%