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ale4655 [162]
3 years ago
10

D over 12 when d= 36

Mathematics
2 answers:
AlekseyPX3 years ago
8 0

the answer would be 36/12, or 3.

Hopefully this helps! Let me know if you have any more questions.

antiseptic1488 [7]3 years ago
6 0

\frac{D}{2} when d=36


You just input the 36 into the variable D.

36/12= 3, so the final answer is three.

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A simple random sample of 90 is drawn from a normally distributed population, and the mean is found to be 138, with a standard d
bagirrra123 [75]

The 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C:  [130.10, 143.90]

<h3>
How to find the confidence interval for population mean from large samples (sample size > 30)?</h3>

Suppose that we have:

  • Sample size n > 30
  • Sample mean = \overline{x}
  • Sample standard deviation = s
  • Population standard deviation = \sigma
  • Level of significance = \alpha

Then the confidence interval is obtained as

  • Case 1: Population standard deviation is known

\overline{x} \pm Z_{\alpha /2}\dfrac{\sigma}{\sqrt{n}}

  • Case 2: Population standard deviation is unknown.

\overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}

For this case, we're given that:

  • Sample size n = 90 > 30
  • Sample mean = \overline{x} = 138
  • Sample standard deviation = s = 34
  • Level of significance = \alpha = 100% - confidence = 100% - 90% = 10% = 0.1 (converted percent to decimal).

At this level of significance, the critical value of Z is: Z_{0.1/2} = ±1.645

Thus, we get:

CI = \overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}\\CI = 138 \pm 1.645\times \dfrac{34}{\sqrt{90}}\\\\CI \approx 138 \pm 5.896\\CI \approx [138 - 5.896, 138 + 5.896]\\CI \approx [132.104, 143.896] \approx [130.10, 143.90]

Thus, the 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C:  [130.10, 143.90]

Learn more about confidence interval for population mean from large samples here:

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