Evaluate the expression 12.6 (2-3) + 2^6 divided by 4

1 answer:

7 0

$\bf 12.6(2-3)+2^6 \div 4\implies \stackrel{\mathbb{P~E~M~D~A~S}}{12.6(\stackrel{\stackrel{2-3}{\downarrow }}{-1})}+2^6\div 4\implies 12.6(-1)+\stackrel{\stackrel{2^6}{\downarrow }}{64}\div 4 \\\\\\ \stackrel{\stackrel{12.6(-1)}{\downarrow }}{-12.6}+64\div 4\implies -12.6+\stackrel{\stackrel{64\div 4}{\downarrow }}{16}\implies \stackrel{\stackrel{-12.6+16}{\downarrow }}{3.4}$

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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

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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 = $\overline{x}$
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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Step-by-step explanation:

Let $\mu$ be the population mean .