Mean, Standard Deviation & Variance

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

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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.

$\displaystyle \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: $\displaystyle \mu = \frac{\sum fx}{\sum f} = \frac{396}{37} = 10.7$

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

**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.

$\displaystyle \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}$$\displaystyle \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: $\displaystyle \mu = \frac{\sum fx}{\sum f} = \frac{3917.5}{120} = 32.6$

Variance: This went terribally bad as I got a negative answer which is impossible. :o

**The real answer:**

Mean: 32.7 (My answer incorrect by 0.1 (Doh))

Variance: 162.03

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Please could someone help? Thank you in advance. :)