Question

According to Gallup’s annual poll on personal finances, while most U.S. workers reported living comfortably now, many expected a downturn in their lifestyle when they stop working. Approximately 55% said they have enough money to live comfortably now and expected to do so in the future. If you select a sample of 200 U.S. workers, the probability is 90% that less than what sample percentage will say they expect to live comfortably? Answer with the percentage in decimal form to 4 decimal places.

Answer 1

- The required percentage to 4 d.p. = 59.5126%

- The required percentage in decimal point to 4 d.p. = 0.5951

Explanation:

Proportion that said they have enough money to live comfortably now and expected to do so in the future = 55% = 0.55

Sample size = n = 200

The Central Limit Theorem ensures that we can say that:

1) The sample proportion is equal to the population proportion.

p = μₓ = μ = 0.55

2) Standard Deviation of the sampling distribution = σₓ = √[p(1-p)/n] = √(0.55×0.45/200) = 0.0352

3) We can say that the sample distribution approximates a normal distribution especially when

np > 5 and nq > 5 (which is true for this sample size)

The probability is 90% that less than what sample percentage will say they expect to live comfortably

To find this sample percentage, let that that sample percentage be x' and its z-score be z'

z' = (x' - μ)/σₓ

But we are told that

P(z < z') = 90% = 0.90

Using the normal distribution table

z' = 1.282

1.282 = (x' - 0.55)/0.0352

x' = 0.55 + (0.0352×1.282) = 0.5951264

Hence, the required sample percentage = 59.5126% to 4 d.p.

Hope this Helps!!!