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Pavel [41]
3 years ago
13

A school has sixth-grade students that are going on a field trip at the end of the school year. the students will be divided int

o five groups, each group riding on a different bus. there are a total number of 242 students going on the field trip. write and solve an equation to determine s, the number of students that will ride on both bus #4.
buses number of students
bus 1 62
bus 2 64
bus 3 55
bus 4 s
Mathematics
1 answer:
Nikitich [7]3 years ago
4 0

Answer:

61

Step-by-step explanation:

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Step-by-step explanation:

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a) \frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

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

19.08 \leq \sigma \leq 66.56

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

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}}

On this case we need to find the sample standard deviation with the following formula:

s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}

The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=9-1=8

The Confidence interval is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and the critical values are:

\chi^2_{\alpha/2}=20.09

\chi^2_{1- \alpha/2}=1.65

And replacing into the formula for the interval we got:

\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

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

19.08 \leq \sigma \leq 66.56

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

4 0
4 years ago
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