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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 =

- Sample standard deviation = s
- Population standard deviation =

- Level of significance =

Then the confidence interval is obtained as
- Case 1: Population standard deviation is known

- Case 2: Population standard deviation is unknown.

For this case, we're given that:
- Sample size n = 90 > 30
- Sample mean =
= 138 - Sample standard deviation = s = 34
- Level of significance =
= 100% - confidence = 100% - 90% = 10% = 0.1 (converted percent to decimal).
At this level of significance, the critical value of Z is:
= ±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]](https://tex.z-dn.net/?f=CI%20%3D%20%5Coverline%7Bx%7D%20%5Cpm%20Z_%7B%5Calpha%20%2F2%7D%5Cdfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%5C%5CCI%20%3D%20138%20%5Cpm%201.645%5Ctimes%20%5Cdfrac%7B34%7D%7B%5Csqrt%7B90%7D%7D%5C%5C%5C%5CCI%20%5Capprox%20138%20%5Cpm%205.896%5C%5CCI%20%5Capprox%20%5B138%20-%205.896%2C%20138%20%2B%205.896%5D%5C%5CCI%20%5Capprox%20%5B132.104%2C%20143.896%5D%20%5Capprox%20%5B130.10%2C%20143.90%5D)
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:
brainly.com/question/13770164
Answer:
D
Step-by-step explanation:
379.00 - 15.00 + 154.00 x 3 + 65.00 + 13.00 = 904
0.55 x 63 = 34.65
34.65 + 904 = 938.65
5x+8=-27
8 more means addition. -27 would be total. 5 times the number ( in this case x) would mean multiplication.
Answer:
B) 47
Step-by-step explanation:
Given:
N = linear distance between the armpits in cm
2.5P = 2N
or
P = (2 ÷ 2.5)N
or
P = 0.8N ................(1)
also,
N = P + 10,
substituting the value of P from (1) we get
N = 0.8N + 10
or
N - 0.8N = 10
or
0.2N = 10
or
N = 10 ÷ 0.2
or
N = 50
Hence
the value is closest to option (B) 47