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2025-MS1004 practice quiz 6 -- optional homework -- for self-assessment

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Calculation: Let ¯ X denote the average of 30 observations. Then ¯ X follows a normal distribution too, with a mean of μ=40 and a standard deviation σ √ n = 8 √ 30 ≈1.4606. So the standardisation of ¯ X yields P( ¯ X <37)=P( ¯ X −40 1.4606 < 37−40 1.4606 )=P(Z<−2.05)=P(Z>2.05)=0.02018.   Select "Continue" to continue.

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The question presents a calculation involving the sampling distribution of the sample mean X̄ for 30 observations with μ = 40 and σ/√n = 8/√30 ≈ 1.4606. It then evaluates P(X̄ < 37) by standardizing to Z: (X̄ − 40) / 1.4606 < (37 − 40) / 1.4606 = −3 / 1.4606 ≈ −2.05. This co......Login to view full explanation

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