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Seventy percent of adults favor some kind of government control on the prices of medicines. What is the probability that the proportion of adults in a random sample of 400 who favor some kind of government control is greater than 0.76?

a. 0.996
b. 0.0044
c. 0.0229
d. 0.005

User Etheryte
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Final answer:

Using normal approximation, we calculate the probability of more than 76% of a sample of 400 adults favoring government control on medicine prices. The probability is 0.0072, which does not match any of the provided answer choices.

Step-by-step explanation:

The question is asking to find the probability that in a sample of 400 adults, more than 76% (0.76) would favor government control on the prices of medicines, given that 70% of the adult population favors it. To solve this problem, we can use normal approximation to the binomial distribution since the sample size is large.

First, we find the mean (μ) and standard deviation (σ) for the sample proportion:

  • μ = p = 0.70
  • σ = sqrt((p(1-p))/n) = sqrt((0.70(0.30))/400) = 0.0245

Next, we standardize the sample proportion using the z-score formula:

z = (x - μ) / σ

Plugging in the values, we get:

z = (0.76 - 0.70) / 0.0245 = 2.449

Using the z-table, we find the probability corresponding to z = 2.449, which is approximately 0.9928. This is the probability of getting a proportion of 0.76 or less. However, we need the probability of getting more than 0.76. So we subtract this value from 1:

P(X > 0.76) = 1 - P(X <= 0.76) = 1 - 0.9928 = 0.0072

Thus, none of the given options (a. 0.996, b. 0.0044, c. 0.0229, d. 0.005) exactly match our calculated probability. If this were a multiple-choice question, we'd probably need to conclude that there is either a mistake in the question or in the answer choices.

User Raultm
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