Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the flows are generated by waste-water from households and industries, together withsome possible drainage from water stored in the topsoil from previous rainfalls. In a study of an urbansewer system, the following values were obtained for flowrates during dry weather conditions: 423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Assume that the flow rates are normally distributed.

a. Construct a 98% two-sided confidence interval for the standard deviation of the flow rate under dry weather conditions.
b. Explain, only using words, the meaning of the CI you determined in (a).

a)

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

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:

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

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:

The Confidence interval is 0.98 or 98%, the value of and , and the critical values are:

And replacing into the formula for the interval we got:

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

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

percentage - 25%

probability - 0.25

we know,

total amount of students in class = 20

total boys = 10

total girls = 10

no of girls who wear glasses = 10/2 = 5

probability of choosing a girl with glasses is =

number of girls wearing glasses/total number of students in the class = 5/20

thus, percentage of selecting a girl wearing glasses will be

= 5/20 × 100

= 5 × 5

25%

therefore,

probability will be 0.25

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