A camp accrediting association has 2,400 camps in their association. They claim that $45 \%$ of their camps offer specialized programs for individuals with disabilities. Suppose you contact a simple random sample of 200 of the camps and ask them whether they offer specialized programs for individuals with disabilities.

Assuming that the association's claim is correct, what is the approximate probability that between $40 \%$ and $50 \%$ of the camps reply "yes" to your question?

Choose 1 answer:
(A) $P(0.4

Q: A camp accrediting association has 2,400 camps in their association. They claim that $45 \%$ of their camps offer specialized programs for individuals with disabilities. Suppose you contact a simple random sample of 200 of the camps and ask them whether they offer specialized programs for individuals with disabilities. Assuming that the association's claim is correct, what is the approximate probability that between $40 \%$ and $50 \%$ of the camps reply "yes" to your question? Choose 1 answer: (A) $P(0.4<\hat{p}<0.5) \approx 0.69$ (B) $P(0.4<\hat{p}<0.5) \approx 0.74$ (c) $P(0.4<\hat{p}<0.5) \approx 0.79$ () $P(0.4<\hat{p}<0.5) \approx 0.84$

∻Solution by Steps∻ ‖ step 1 ⋮ Identify the given parameters: The population proportion $p = 0.45$, the sample size $n = 200$, and the range of interest $0.4 < \hat{p} < 0.5$. ‖ ‖ step 2 ⋮ Calculate the standard error (SE) of the sample proportion using the formula: $SE = \sqrt{\frac{p(1-p)}{n}}$. Substituting the values, we get $SE = \sqrt{\frac{0.45 \times 0.55}{200}} \approx 0.035$. ‖ ‖ step 3 ⋮ Convert the range of sample proportions to z-scores using the formula: $z = \frac{\hat{p} - p}{SE}$. For $\hat{p} = 0.4$, $z = \frac{0.4 - 0.45}{0.035} \approx -1.43$. For $\hat{p} = 0.5$, $z = \frac{0.5 - 0.45}{0.035} \approx 1.43$. ‖ ‖ step 4 ⋮ Use the standard normal distribution table to find the probabilities corresponding to the z-scores. $P(Z < -1.43) \approx 0.0764$ and $P(Z < 1.43) \approx 0.9236$. ‖ ‖ step 5 ⋮ Calculate the probability that the sample proportion is between 0.4 and 0.5: $P(0.4 < \hat{p} < 0.5) = P(Z < 1.43) - P(Z < -1.43) = 0.9236 - 0.0764 = 0.8472$. ‖ ∻Answer∻ ***D*** ∻Key Concept∻ ⚹Standard Error and Z-scores⚹ ∻Explanation∻ ⚹The standard error measures the variability of the sample proportion, and z-scores help convert sample proportions to standard normal distribution values to find probabilities.⚹◊What is the formula for calculating the approximate probability that between 40% and 50% of the camps reply "yes" to the question?⍭ Generate me a similar question◊

Question

A camp accrediting association has 2,400 camps in their association. They claim that $45 \%$ of their camps offer specialized programs for individuals with disabilities. Suppose you contact a simple random sample of 200 of the camps and ask them whether they offer specialized programs for individuals with disabilities.

Assuming that the association's claim is correct, what is the approximate probability that between $40 \%$ and $50 \%$ of the camps reply "yes" to your question?

Choose 1 answer:
(A) $P(0.4<\hat{p}<0.5) \approx 0.69$
(B) $P(0.4<\hat{p}<0.5) \approx 0.74$
(c) $P(0.4<\hat{p}<0.5) \approx 0.79$
() $P(0.4<\hat{p}<0.5) \approx 0.84$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Identify the given parameters: The population proportion $p = 0.45$, the sample size $n = 200$, and the range of interest $0.4 < \hat{p} < 0.5$. ‖ ‖ step 2 ⋮ Calculate the standard error (SE) of the sample proportion using the formula: $SE = \sqrt{\frac{p(1-p)}{n}}$. Substituting the values, we get $SE = \sqrt{\frac{0.45 imes 0.55}{200}} \approx 0.035$. ‖ ‖ step 3 ⋮ Convert the range of sample proportions to z-scores using the formula: $z = \frac{\hat{p} - p}{SE}$. For $\hat{p} = 0.4$, $z = \frac{0.4 - 0.45}{0.035} \approx -1.43$. For $\hat{p} = 0.5$, $z = \frac{0.5 - 0.45}{0.035} \approx 1.43$. ‖ ‖ step 4 ⋮ Use the standard normal distribution table to find the probabilities corresponding to the z-scores. $P(Z < -1.43) \approx 0.0764$ and $P(Z < 1.43) \approx 0.9236$. ‖ ‖ step 5 ⋮ Calculate the probability that the sample proportion is between 0.4 and 0.5: $P(0.4 < \hat{p} < 0.5) = P(Z < 1.43) - P(Z < -1.43) = 0.9236 - 0.0764 = 0.8472$. ‖ ∻Answer∻ ***D*** ∻Key Concept∻ ⚹Standard Error and Z-scores⚹ ∻Explanation∻ ⚹The standard error measures the variability of the sample proportion, and z-scores help convert sample proportions to standard normal distribution values to find probabilities.⚹◊What is the formula for calculating the approximate probability that between 40% and 50% of the camps reply "yes" to the question?⍭ Generate me a similar question◊

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