Question

Suppose 52%

of politicians are lawyers.

If a random sample of size 436

is selected, what is the probability that the proportion of politicians who are lawyers will be less than 45%

? Round your answer to four decimal places.

Answer 1

Using central limit theorem,

P( < p) = P( Z < - p / sqrt( p (1 - p) / n) )

So ,

P( < 0.45 ) = P( Z < 0.45 - 0.52 / sqrt( 0.52 * 0.48 / 436) )

= P( Z < -2.9256)

= 0.0.0017

Answer 2

To calculate the probability that the proportion of politicians who are lawyers will be less than 45%, we can use the sampling distribution of the sample proportion.

Given: Population proportion (p) = 52% = 0.52 Sample size (n) = 436 Sample proportion (p̂) for the sample proportion being less than 45% = 45% = 0.45

The sampling distribution of the sample proportion (p̂) is approximately normally distributed when the sample size is large, which is the case here (n = 436). The mean (μp̂) of the sampling distribution is equal to the population proportion (p), and the standard deviation (σp̂) is given by:

σp̂ = √[(p * (1 - p)) / n]

Now, let's calculate σp̂:

σp̂ = √[(0.52 * (1 - 0.52)) / 436] σp̂ ≈ √[(0.52 * 0.48) / 436] σp̂ ≈ √[0.2496 / 436] σp̂ ≈ √0.000573394 σp̂ ≈ 0.023939

Now, we want to find the probability that the sample proportion (p̂) is less than 0.45:

Z = (p̂ - p) / σp̂

Z = (0.45 - 0.52) / 0.023939

Z ≈ -2.9318

Now, we need to find the probability corresponding to the Z-score of -2.9318 using a standard normal distribution table or a calculator:

Probability (p) = 0.001972 (rounded to 4 decimal places)

So, the probability that the proportion of politicians who are lawyers will be less than 45% is approximately 0.001972 (or 0.1972%).