Question

An institute reported that 81% of ils members indicate that lack of ethical culture within financial firms has contributed nos! ta the lack of trust in the financial industry. Suppose that you select a sample of 100 institute mierniers. Complete paris (a) through (d) below. Kern at least three decimal places for the standard crrar in order to get an answry clase to the solution in the system. a. What is the probability that the sample percentage indicating that lack of chical culture within financial firms has contributed the most to the lack of trust in the financial industry will be between 52% and 82%? (Type an integer or decimal rounded to four decimal places as needed.) b. The probability is 90% that the sample percentage will be contained within what symmetrical limits of the population percentage? Hints: (1) shade the middle 70% of the curve. (2) charge decimal la percenlages aller finished with calculations. and below % The probability is 90% that the sample percentage will be contained above (Type integers or decimals rounded to one decimal place as needed.) c. The probability is 97% that the sample percentage wil be contained within what symmetrical limits of the population percentage? The probability is 97% that the sample percentage will be contained above (Type integers ar decimals rourdexant decimal place as needed.) and below % d. Suppose you selected a sample of 400 institute members. How does this change your answers in (a) through (y? O A. An increase in simple size increases the standard Tor, causing the value of the answer to part (a) to increase and the ranges in parts (h) and (c) to widen. O B. An increase in sample size decreases the standard error, causing the value of the answer to par (a) to decrease and the ranges in pearls (b) and (c) to narrow. O C. An increase in sample size decreases the standard error, causing the value of the answer to part (a) to increase and the ranges in parte (b) and (c) to narrow. OD. An increase in sample size increases the standard error, causing the value of the answer to part (a) to cecrease and the ranges in parte (b) and (c) to widen.

Answer 1

Given that a population proportion P=0.61 1 sample size n = 100 let 8 = sample proportion Using central limit theorm, Pn Normal (mean = P, std dn= (pere) @ prob [ 0.58 L P L 0.62 ] =pl 0.58-0.6) LP-P 20.62-0.61 7 voon xoss JPCL-e) 0-918034 - P[ -0.651 LZ L 0.2050] = 0.3120 V 6 prob [ P L P L P₂] = 0.90 Using a table I PC -1.645 L z L 1.645] = 0.90 JP[p. 1.645 / PCI-E) L P L P HIGUS / PLED = ogo = P, = P - 1.645 / Pl_po

0-6 ) x 029 loo P = 0.6) - 1.645 60-61x039 I p = 0.61 -0.0802 P = 0.5298 Similarly P2 = 0.61 +0.0802 - = 0.6902 Answer is 50 % & below 69.0% abore @ ke wont Pz & Pq such that PL P L P L P4 J = 0.97 Using a table P [-2.17 LZ 22.11] = 0.97 I P₂ = P-2.17 | PCI-P) & Py = P + 2.12 Jectes = = 0.61 -20170.61X039 a P₂ = 0.61 -0.1058 = 0.5042 I Py = 0.61 +0.1058 = 0.2158 Answering (abor 150.4% & below 71.6% & Answer is @r .) ~ 2.17 0.61 X 0.39 V 100