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Nadya [2.5K]
2 years ago
5

PLS HELPP ASP Whats the sum of 8 1/2 2 1/2

Mathematics
1 answer:
kompoz [17]2 years ago
8 0

Answer:

11

Step-by-step explanation:

1/2 + 1/2 is 1

8 + 2 = 10

1 + 10 = 11

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83 random samples were selected from a normally distributed population and were found to have a mean of 32.1 and a standard devi
arlik [135]

Answer:

\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}

4.525 \leq \sigma^2 \leq 7.602

Now we just take square root on both sides of the interval and we got:

2.127 \leq \sigma \leq 2.757

Step-by-step explanation:

Information given

\bar X=32.1 represent the sample mean

\mu population mean (variable of interest)

s=2.4 represent the sample standard deviation

n=83 represent the sample size  

Confidence interval

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The degrees of freedom given by:

df=n-1=8-1=7

The confidence level is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,82)" "=CHISQ.INV(0.95,82)". so for this case the critical values are:

\chi^2_{\alpha/2}=104.139

\chi^2_{1- \alpha/2}=62.132

The confidence interval is given by:

\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}

4.525 \leq \sigma^2 \leq 7.602

Now we just take square root on both sides of the interval and we got:

2.127 \leq \sigma \leq 2.757

5 0
3 years ago
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