Mean, Standard Deviation & Variance

**Simplicity** · December 30th 2007, 06:07 AM

I have 2 question which I'm stuck on. I get my $f$ and $x$ mixed up, I believe?

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Q1: The following table shows the number of take-away meals eaten by a random sample of students during a particular month.

$\begin{tabular}{|l|l|l|l|l|l|l|l|} \hline Number of meals & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline Number of students & 2 & 5 & 12 & 8 & 5 & 3 & 2 \\ \hline \end{tabular}$

Calculate an estimate of the mean and an estimate of the standard deviation of the numbers of take-away meals eaten by the population from which this sample was taken.

My attempt:

Mean: $\mu = \frac{\sum fx}{\sum f} = \frac{396}{37} = 10.7$
Standard Deviation: $Var = \frac{\sum fx^2}{\sum f} - \mu^2 = \frac{4320}{37} - (10.7)^2 = 2.21 \implies S.D = \sqrt{Var} = \sqrt{2.21} = 1.49$

The real answer:
Mean: 10.7 (My Answer Correct)
Standard Deviation: 1.51 (My Answer Incorrect

)

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Q2: Whig and Penn, solicitors, monitor the time spent on consultations with a random sample of 120 of their clients. The times, to the nearest minutes, are summarised in the following table.

$\begin{tabular}{|l|l|l|l|l|l|} \hline Time & 10-14 & 15-19 & 20-24 & 25-29 & 30-34 \\ \hline Number of clients & 2 & 5 & 17 & 33 & 27 \\ \hline \end{tabular}$ $\begin{tabular}{l|l|l|l|} \hline 35-44 & 45-59 & 60-89 & 90-119 \\ \hline 25 & 7 & 3 & 1 \\ \hline \end{tabular}$

Calculate estimates of the mean and variance of the population of times from which these data were obtained.

My attempt:

Mean: $\mu = \frac{\sum fx}{\sum f} = \frac{3917.5}{120} = 32.6$
Variance: This went terribally bad as I got a negative answer which is impossible.

The real answer:
Mean: 32.7 (My answer incorrect by 0.1

)
Variance: 162.03

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Please could someone help? Thank you in advance.

**badgerigar** · December 30th 2007, 03:06 PM

When you are getting such small errors in your answers, it is likely that they result simply from rounding at different places in earlier steps in the arithmetic. I can't see any mistakes in your method for any of the questions.

You are right that Variance cannot be negative. The method you used for the first question should get you the right answer.

Good job. You appear to know what you are doing.

**mathgeek777** · December 30th 2007, 04:32 PM

I just did the same work you did on the standard deviation in the first problem and got a slightly different answer.

The answer before taking the square root (unrounded) is 2.266756757 and the variance (unrounded) is 1.505575225, which would naturally round off to 1.51 and thus match the book answer.

I would agree with badgerigar's reasoning if it were not for the fact that I got 2.26 and you got 2.21 before taking the square root. Comparing the two answers, I seriously doubt there was some sort of rounding issue in the problem solving.

It is quite probable that you simply typed it into your calculator incorrectly. Try typing the problem in again, making sure that the parentheses are in the correct locations. You should be able to get the correct answer.

EDIT: Be careful with the rounding. You will have to take it out to 4 decimal places at the minimum in order to get the correct answer when taking the square root.

**Simplicity** · December 31st 2007, 02:29 AM

For Q1, I consider 'Number of Meals' as $x$ and 'Number of Students' as $f$ , Am I correct? Also, when finding the variance as $ Var = \frac{\sum fx^2}{\sum f} - \mu^2$ , do I just square $x^2$ or is it $(fx)^2$ ?

For Q2, I consider mid 'Time' as $x$ and 'Number of Clients' as $f$ , is this correct?

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