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suppose 46% of politicians are lawyers. if a random sample of size 662 is selected, what is the probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by more than 4% ? round your answer to four decimal places.

User Vexatus
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The probability that the proportion of politicians who are lawyers will differ from the total politician proportion by more than 4% is approximately 0.0294 (rounded to four decimal places).

To solve this problem, we can use the normal approximation to the binomial distribution since the sample size is reasonably large. The standard deviation of the sampling distribution of the sample proportion (p) is given by the formula:

σ_p = √p⋅(1−p)/n

where

p is the population proportion (proportion of politicians who are lawyers), and n is the sample size.

In this case,

p=0.46 and n=662. Plugging in these values:

σ_p = √0.46⋅(1−0.46)/ 662

​Now, we want to find the probability that the sample proportion differs from the population proportion by more than 4%. This means we're interested in finding P(∣ p −p∣>0.04), where p is the sample proportion.

To do this, we need to convert the difference to a z-score using the formula:

z= ∣ p −p∣/ σ_p

​And then find the probability of ∣z∣ being greater than a certain value.

P(∣z∣>some value)

Finally, we round the answer to four decimal places.

Let's perform the calculations:

σ_p = √0.46⋅(1−0.46)/ 662

σ_p ≈0.0213

Now, calculate the z-score:

z= 0.04/ 0.0213

z≈1.8797

Now, find the probability:

P(∣z∣>1.8797)

You can use a standard normal distribution table, calculator, or software to find this probability. For a z-score of 1.8797, the probability is approximately 0.0294.

So, the probability that the proportion of politicians who are lawyers will differ from the total politician proportion by more than 4% is approximately 0.0294 (rounded to four decimal places).

User Holtavolt
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