Question

An institute reported that

6262​%

of its members indicate that lack of ethical culture within financial firms has contributed most to the lack of trust in the financial industry. Suppose that you select a sample of 100 institute members. Complete parts ​(a) through ​(d) below.

a. What is the probability that the sample percentage indicating that lack of ethical culture within financial firms has contributed the most to the lack of trust in the financial industry will be between

6060​%

and

6767​%?

. 5083.5083

​(Type an integer or decimal rounded to four decimal places as​ needed.)b. The probability is

6060​%

that the sample percentage will be contained within what symmetrical limits of the population​ percentage?The probability is

6060​%

that the sample percentage will be contained above

nothing ​%

and below

nothing ​%.

Answer 1

Solution:

Given: An institute reported that 62​% of its members indicate that lack of ethical culture within financial firms has contributed most to the lack of trust in the financial industry.

That is: p = 0.62

Sample size = 100

Part a) What is the probability that the sample percentage indicating that lack of ethical culture within financial firms has contributed the most to the lack of trust in the financial industry will be between 60​% and 67​%?

That is find:

Since sample size n = 100 is large and

and

hence , sampling distribution of sample proportions is approximately Normal with mean of sample proportions is:

and standard deviation of sample proportion is:

Find z score for 0.60 and 0.67:

and

Thus

Look in z table for z = 1.0 and 0.03 as well as z = -0.4 and 0.01 and find corresponding area.

P( Z < -0.41 ) = 0.3409

P( Z < 1.03) = 0.8485

Thus

Part b)  The probability is 60​% that the sample percentage will be contained within what symmetrical limits of the population​ percentage?

First find z value such that:

P( z₁ < Z < z₂) = 0.60

Area in two tails = 1 - 0.60 = 0.40

Thus area in left tail = 0.40 / 2 = 0.20

Thus look in z table for area = 0.2000 or its closest area and find corresponding z value.

Area 0.2005 is closest to 0.2000 and it corresponds to -0.8 and 0.04

Thus z = -0.84

Thus z value for right tailed 0.2000 area is 0.84.

Thus we use following formula to find sample proportions:

and

Thus the probability is 60​% that the sample percentage will be contained above 57.92 ​% and below 66.08 ​%.