Questions: Per Capita Income of Delaware Residents In a recent year, Delaware had the highest per capita annual income with 51,803. If σ=4850, answer the following questions. Assume that the sample is taken from a large population and the correction factor can be ignored. Round intermediate z-value calculations to two decimal places and the final answers to at least four decimal places. Part 1 of 2 What is the probability that a random sample of 36 state residents had a mean income greater than 50,200 ? P(X̅>50,200)=.976 Part: 1 / 2 Part 2 of 2 What is the probability that a random sample of 36 state residents had a mean income less than 48,800 ? P(X̅<48,800)=

Per Capita Income of Delaware Residents In a recent year, Delaware had the highest per capita annual income with 51,803. If σ=4850, answer the following questions. Assume that the sample is taken from a large population and the correction factor can be ignored. Round intermediate z-value calculations to two decimal places and the final answers to at least four decimal places.

Part 1 of 2

What is the probability that a random sample of 36 state residents had a mean income greater than 50,200 ?
P(X̅>50,200)=.976

Part: 1 / 2

Part 2 of 2

What is the probability that a random sample of 36 state residents had a mean income less than 48,800 ?
P(X̅<48,800)=
Transcript text: Per Capita Income of Delaware Residents In a recent year, Delaware had the highest per capita annual income with $\$ 51,803$. If $\sigma=\$ 4850$, answer the following questions. Assume that the sample is taken from a large population and the correction factor can be ignored. Round intermediate $z$-value calculations to two decimal places and the final answers to at least four decimal places. Part 1 of 2 What is the probability that a random sample of 36 state residents had a mean income greater than $\$ 50,200$ ? \[ P(\bar{X}>50,200)=.976 \] $\square$ Part: $1 / 2$ Part 2 of 2 What is the probability that a random sample of 36 state residents had a mean income less than $\$ 48,800$ ? \[ P(\bar{X}<48,800)= \] $\square$
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Solution

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Solution Steps

Step 1: Calculate the Standard Error (SE)

The standard error (SE) of the mean is calculated using the formula: $$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$$ Substituting the given values: $$\sigma_{\bar{X}} = \frac{4850}{\sqrt{36}} = 808.333$$

Step 2: Calculate the Z-score

The Z-score is calculated using the formula: $$Z = \frac{\bar{X} - \mu}{\sigma_{\bar{X}}}$$ Substituting the given values: $$Z = \frac{50200 - 51803}{808.333} = -1.983$$

Step 3: Find the Probability

Since we are looking for \(P(\bar{X} > k)\), we use \(1 - P(Z)\): Substituting the Z-score into the standard normal distribution: $$P = 0.976$$

Final Answer:

The probability that the sample mean is greater than 50200 is 0.976.

Step 1: Calculate the Standard Error (SE)

The standard error (SE) of the mean is calculated using the formula: $$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$$ Substituting the given values: $$\sigma_{\bar{X}} = \frac{4850}{\sqrt{36}} = 808.333$$

Step 2: Calculate the Z-score

The Z-score is calculated using the formula: $$Z = \frac{\bar{X} - \mu}{\sigma_{\bar{X}}}$$ Substituting the given values: $$Z = \frac{48800 - 51803}{808.333} = -3.715$$

Step 3: Find the Probability

Since we are looking for \(P(\bar{X} < k)\), we use \(P(Z)\): Substituting the Z-score into the standard normal distribution: $$P = 0.0001$$

Final Answer:

The probability that the sample mean is less than 48800 is 0.0001.

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