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determine the .05 and .95 quantities for the weight of a single hamburger made by jon to determine the middle 90% probability interval for a normal random variable

User Wuarmin
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Answer:

So, the middle 90% probability interval for the weight of a single hamburger is approximately \(167.1\) grams to \(232.9\) grams.

Explanation:

Certainly, let's calculate the values using the given information:

Given:

- Mean (\(\mu\)): 200 grams

- Standard Deviation (\(\sigma\)): 20 grams

- Z-score for .05 quantile: -1.645

- Z-score for .95 quantile: 1.645

Substitute these values into the formulas:

\[ \text{Lower bound} = 200 - 1.645 \times 20 \]

\[ \text{Upper bound} = 200 + 1.645 \times 20 \]

Calculating:

\[ \text{Lower bound} = 200 - (1.645 \times 20) = 200 - 32.9 = 167.1 \]

\[ \text{Upper bound} = 200 + (1.645 \times 20) = 200 + 32.9 = 232.9 \]

So, the middle 90% probability interval for the weight of a single hamburger is approximately \(167.1\) grams to \(232.9\) grams.

User Alphanumeric
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