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Chapter 9: Problem 482

Find the variance of the sample of observations \(2,5,7,9,12\).

Short Answer

Expert verified
The variance of the sample of observations \(2, 5, 7, 9, 12\) is \(11.6\).

Step by step solution

01

Calculate the mean of the sample.

First, we need to find the mean of the sample. The mean is the average of all the observations. To find the mean, add all the observations and divide the sum by the number of observations. Mean = \(\frac{sum \ of \ observations}{number \ of \ observations}\) The sum of the observations: \(2 + 5 + 7 + 9 + 12 = 35\) Number of observations: \(5\) Mean = \(\frac{35}{5} = 7\)
02

Find the differences between each observation and the mean.

Now, we need to find the differences between each observation and the mean calculated in Step 1. Difference 1: \(2 - 7 = -5\) Difference 2: \(5 -7 = -2\) Difference 3: \(7- 7 = 0\) Difference 4: \(9 - 7 = 2\) Difference 5: \(12 - 7 = 5\)
03

Square the differences.

In this step, we need to square each of the differences found in Step 2. Squared Difference 1: \((-5)^2 = 25\) Squared Difference 2: \((-2)^2 = 4\) Squared Difference 3: \((0)^2 = 0\) Squared Difference 4: \((2)^2 = 4\) Squared Difference 5: \((5)^2 = 25\)
04

Find the mean of the squared differences.

To find the mean of the squared differences, add all the squared differences and divide the sum by the number of squared differences. Mean of Squared Differences = \(\frac{sum \ of \ squared \ differences}{number \ of \ squared \ differences}\) The sum of squared differences: \(25 + 4 + 0 + 4 + 25 = 58\) Number of squared differences: \(5\) Mean of Squared Differences = \(\frac{58}{5} = 11.6\)
05

The variance is the result obtained in step 4.

Finally, the variance of the sample of observations is the result obtained in step 4, which is 11.6. So, the variance of the sample of observations \(2,5,7,9,12\) is \(11.6\).

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Most popular questions from this chapter

My wife wanted to know whether putting cut flowers into a certain chemical solution (we'll call it 'Flower-Life') would prolong their life, so we designed the following experiment. She bought 2 fresh blooms of 25 different kinds of flowers \(-2\) roses, 2 irises, 2 carnations, and so on We then put one of each pair in a vase of water, and their partners in a vase containing 'Flower-Life'. Both vases were put side by side in the same room, and the length of life of each flower was noted. We then had 2 matched samples, so the results could be tested for significance by Wilcoxon's Signed Ranks Test. This revealed a smaller rank total of \(50 .\) Is there a statistical difference between 'Flower-Life' and plain water?

Suppose the random vector \((\mathrm{X}, \mathrm{Y})\) is distributed with probability density, \(\mathrm{f}(\mathrm{x}, \mathrm{y})=\mathrm{x}+\mathrm{y}\) \(0<\mathrm{x}<1\) and \(=0 \quad 0<\mathrm{y}<1\) otherwise. Find \(E[X Y], E[X+Y]\) and \(E(X)\).

Given that \(\mathrm{x}\) has a normal distribution with mean 10 and standard deviation 4, find \(\mathrm{P}(\mathrm{x}<15)\).

A plant manager claims that on the average no more than 5 service calls per hour are made by the plant's workers. Suppose in one particular hour, 9 service calls were required. At a \(5 \%\) level of significance, could we now reject the plant manager's claim?

Let \(\mathrm{x}_{1}, \ldots \ldots, \mathrm{x}_{\mathrm{n}}\) be a random sample from a distribution having mean \(\mu\). Show that the sample mean, \(\overline{\mathrm{x}}={ }^{\mathrm{n}} \sum_{\mathrm{i}=1} \mathrm{x}_{\mathrm{i}}\) is an unbiased estimator of the population mean.
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