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Chapter 23: Problem 1

Find the variance and standard deviation of the following sets of data: (a) 611109789 (b) \(\begin{array}{lllll}5.3 & 7.2 & 9.1 & 8.6 & 5.9 & 7.3\end{array}\) (c) \(-6-6-5-10 &{2} &{1} &{0} &{-2} \end{array}\) Which set has the greatest variation?

Short Answer

Expert verified
Answer: Set C has the greatest variation.

Step by step solution

01

Find the mean of the data

To find the mean, sum all the data points and divide by the number of data points. In this case, the mean is \(\bar{x} = \frac{6+1+1+0+9+7+8+9}{8} = \frac{41}{8} = 5.125\).
02

Calculate the variance

Now we will plug the data points and the mean into the variance formula and calculate \(\sigma^2\): \(\sigma^2 = \frac{1}{8} ((6-5.125)^2 + (1-5.125)^2 + (1-5.125)^2 + (0-5.125)^2 + (9-5.125)^2 + (7-5.125)^2 + (8-5.125)^2 + (9-5.125)^2) = \frac{1}{8}(0.765625+16.765625+16.765625+26.390625+14.890625+3.515625+8.265625+14.890625) \approx 12.859\).
03

Calculate the standard deviation

The standard deviation is the square root of the variance: \(\sigma = \sqrt{12.859} \approx 3.59\). #(b)#
04

Find the mean of the data

To find the mean, sum all the data points and divide by the number of data points. In this case, the mean is \(\bar{x} = \frac{5.3+7.2+9.1+8.6+5.9+7.3}{6} = \frac{43.4}{6} \approx 7.23\).
05

Calculate the variance

Now we will plug the data points and the mean into the variance formula and calculate \(\sigma^2\): \(\sigma^2 = \frac{1}{6}((5.3-7.23)^2+(7.2-7.23)^2+(9.1-7.23)^2+(8.6-7.23)^2+(5.9-7.23)^2+(7.3-7.23)^2) = \frac{1}{6}(3.74+0.001+3.44+1.86+1.76+0.005) \approx 1.63\).
06

Calculate the standard deviation

The standard deviation is the square root of the variance: \(\sigma = \sqrt{1.63} \approx 1.28\). #(c)#
07

Find the mean of the data

To find the mean, sum all the data points and divide by the number of data points. In this case, the mean is \(\bar{x} = \frac{-6-6+5+10+2+1+0+-2}{8} = \frac{4}{8} = 0.5\).
08

Calculate the variance

Now we will plug the data points and the mean into the variance formula and calculate \(\sigma^2\): \(\sigma^2 = \frac{1}{8}((-6-0.5)^2+(-6-0.5)^2+(5-0.5)^2+(10-0.5)^2+(2-0.5)^2+(1-0.5)^2+(0-0.5)^2+(-2-0.5)^2) = \frac{1}{8}(42.25+42.25+20.25+90.25+2.25+0.25+0.25+6.25) \approx 25.50\).
09

Calculate the standard deviation

The standard deviation is the square root of the variance: \(\sigma = \sqrt{25.50} \approx 5.05\).
10

Determine which set has the greatest variation

Comparing standard deviations, we can see that set C has the greatest variation (standard deviation = 5.05).

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Most popular questions from this chapter

Studies of a particular type of car tyre show that the mileage for which it can be used legally follows a normal distribution with mean 38000 miles and standard deviation 2500 miles. The manufacturers claim that ' 9 out of 10 of our tyres last more than 35000 miles'. Is the claim justified?

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