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Is the below solution correct....because the probability of obtaining a sample which would have the mean height of 70 will fall under the intervals calculated using the mean+/- 3(SD) [99% Confidence level]

Sample problem: In general, the mean height of women is 65″ with a standard deviation of 3.5″. What is the probability of finding a random sample of 50 women with a mean height of 70″, assuming the heights are normally distributed?

z = (x – μ) / (σ / √n) = (70 – 65) / (3.5/√50) = 5 / 0.495 = 10.1

The key here is that we’re dealing with a sampling distribution of means, so we know we have to include the standard error in the formula. We also know that 99% of values fall within 3 standard deviations from the mean in a normal probability distribution (see 68 95 99.7 rule). Therefore, there’s less than 1% probability that any sample of women will have a mean height of 70″.

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  • $\begingroup$ 1. What is the question? 2. You should improve a detail about the language. I guess that "with a mean height of 70" should be "with a mean height of 70 or larger" $\endgroup$ May 23, 2019 at 7:05
  • $\begingroup$ I meant to clarify the last sentence.... about the probability of obtaining a sample which would have the mean height of 70..why is the probability of obtaining such a sample less than 1% since it will fall under the intervals calculated using the mean+/- 3(SD) $\endgroup$
    – Udit Goel
    May 23, 2019 at 7:52
  • $\begingroup$ Did you calculate mean +/- 3(SD) and what is the result? $\endgroup$ May 23, 2019 at 9:47
  • $\begingroup$ The interval will be 3*3.5+/- 65 which gives me 75.5/54.5...which means that 99.7 percent values (approx) will lie between 54.5 to 75.5.. please do correct me if my interpretation is wrong. $\endgroup$
    – Udit Goel
    May 23, 2019 at 12:58
  • $\begingroup$ "The key here is that we’re dealing with a sampling distribution of means" The distribution of the means of samples has a different variance/sd than the population distribution. $\endgroup$ May 24, 2019 at 6:50

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