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deadman7628
26.07.2019 •
Mathematics
Incomes in a certain town are strongly right-skewed with mean $36,000 and standard deviation $7000. a random sample of 75 households is taken. what is the probability the sample mean is greater than $37,000? 0.1075 0 0.4432
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Ответ:
First, calculate the z-score:
z = (x - μ) / σ = (37000 - 36000) / 7000 = 0.143
The probability P(x > 37000$) = 1 - P(x < 37000$),
therefore we need to look up at a normal distribution table in order to find
P(z < 0.143) = 0.55567
And
P(x > 37000$) = 1 - 0.55567 = 0.44433
Hence, there is a 44.4% probability that the sample mean is greater than $37,000.
Ответ: