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Math  /  Data & Statistics

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Coffee: The National Coffee Association reported that 65%65 \% of U.S. adults drink coffee daily. A random sample of 300U.S300 \mathrm{U} . \mathrm{S}. adults is selected. Round your answers to at least four decimal places as needed.
Part 1 of 6 (a) Find the mean μp\mu_{p}.
The mean μp^\mu_{\hat{p}} is 0.65 .
Part 2 of 6 (b) Find the standard deviation σp^\sigma_{\hat{p}}.
The standard deviation σp^\sigma_{\hat{p}} is 0.0275 .
Part 3 of 6 (c) Find the probability that more than 66%66 \% of the sampled adults drink coffee daily.
The probability that more than 66%66 \% of the sampled adults drink coffee daily is 0.3564 .
Part 4 of 6 (d) Find the probability that the proportion of the sampled adults who drink coffee daily is between 0.57 and 0.71 .
The probability that the proportion of the sampled adults who drink coffee daily is between 0.57 and 0.71 is 0.9836 . Part 5 of 6 Cava For 1 ater Submi (e) Find the probability that less than 60%60 \% of sampled adults drink coffee daily.
The probability that less than 60%60 \% of sampled aduts drink coffee đaity is □\square .

Studdy Solution

STEP 1

1. The sample size n=300 n = 300 .
2. The population proportion p=0.65 p = 0.65 .
3. The sampling distribution of the sample proportion p^\hat{p} is approximately normal because np np and n(1−p) n(1-p) are both greater than 5.

STEP 2

1. Calculate the mean μp^\mu_{\hat{p}}.
2. Calculate the standard deviation σp^\sigma_{\hat{p}}.
3. Use the normal distribution to find the probability for different scenarios.

STEP 3

The mean of the sampling distribution of the sample proportion p^\hat{p} is given by:
μp^=p=0.65\mu_{\hat{p}} = p = 0.65

STEP 4

The standard deviation of the sampling distribution of the sample proportion p^\hat{p} is given by:
σp^=p(1−p)n=0.65×0.35300\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.65 \times 0.35}{300}}
Calculate σp^\sigma_{\hat{p}}:
σp^=0.2275300≈0.0275\sigma_{\hat{p}} = \sqrt{\frac{0.2275}{300}} \approx 0.0275

STEP 5

To find the probability that less than 60%60\% of sampled adults drink coffee daily, we need to calculate:
P(p^<0.60)P(\hat{p} < 0.60)
Calculate the z-score for p^=0.60\hat{p} = 0.60:
z=0.60−0.650.0275≈−0.050.0275≈−1.8182z = \frac{0.60 - 0.65}{0.0275} \approx \frac{-0.05}{0.0275} \approx -1.8182
Use the standard normal distribution to find P(Z<−1.8182)P(Z < -1.8182).
Using a standard normal distribution table or calculator, find:
P(Z<−1.8182)≈0.0344P(Z < -1.8182) \approx 0.0344
The probability that less than 60%60\% of sampled adults drink coffee daily is approximately:
0.0344 \boxed{0.0344}

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