Question on mean and standard deviation.

In summary: To answer your other question, we don't "need" to find the value of t. In fact, it looks like the value of t doesn't actually matter here because the sum of the time measurements has been given to us. However, it is possible to find t if you want. Now, the sum of the time measurements is:\sum_{i=1}^{12}t_i = (12)(35) - 15 = 420 - 15 = 405 The value of t would be the only number such that t1 + t2 + t3 + ... + t10 + t11 + t12 = 405. That number is the arithmetic mean of t1, t2, t3, ...,
  • #1
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The length of time, t minutes, taken to do the crossword in a certain newspaper was observed on
12 occasions. The results are summarised below.
Σ(t − 35) = −15
Σ(t − 35)^2 = 82.23
Calculate the mean and standard deviation of these times taken to do the crossword.

My real problem is finding out the value of t. or can we find the answer without finding t??. I took t-35 as y, and hence got y's mean as -1.25. Then I am stuck.
I can't make head or tail of the answer provided for the mean:( why do we deduct 35 -1.25?)

Answer
mean = 35 – 15/12
= 33.75 (33.8) minutes

sd = 82.23/12 ( 15/12)
= 2.3 minutes
 
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  • #2
You took 12 time measurements, and you'll label them:

t1, t2, t3, ... ,t10, t11, t12.

By definition, the arithmetic mean of these time measurements is just: [tex] \bar{t} = \frac{t_1 + t_2 + t_3 + \ldots + t_{10} + t_{11} + t_{12}}{12} = \frac{1}{12}\sum_{i=1}^{12} t_i [/tex]But you haven't been given [itex] \sum_{i=1}^{12} t_i [/itex]. In this problem, what you've been given is that[tex]\sum_{i=1}^{12} (t_i - 35) = -15[/tex]We can write this as:[tex] \sum_{i=1}^{12}t_i - \sum_{i=1}^{12} 35 = \sum_{i=1}^{12}t_i -(12)(35) = -15[/tex][tex]\sum_{i=1}^{12}t_i = (12)(35) - 15 [/tex]Therefore:[tex]\bar{t} = \frac{1}{12}\sum_{i=1}^{12}t_i = \frac{(12)(35)}{12} - \frac{15}{12} = 35 - \frac{15}{12}[/tex]
 
  • #3
I should point out that your method was fine too (and totally equivalent). You said, let yi = ti - 35 so that:[tex]\bar{y} = \frac{1}{12}\sum_{i=0}^{12}y_i = -\frac{15}{12} [/tex]Now, if yi = ti - 35, then it follows that ti = 35 + yi. It also follows that [itex] \bar{t} = 35 + \bar{y}[/itex]. Hence[tex] \bar{t} = 35 - \frac{15}{12} [/tex]My only assumption here was that if you apply a constant offset to all your measurements, their mean will also differ from the mean of the original measurements by that constant offset. This is pretty easy to prove just using the definition of a mean (EDIT: in fact, you can take the steps in my previous post as being a proof of that).
 

What is the purpose of calculating mean and standard deviation?

The mean and standard deviation are two important statistical measures that help to summarize and understand a set of data. The mean, also known as the average, gives an idea of the central tendency of the data, while the standard deviation measures the variability or spread of the data points around the mean.

How do you calculate the mean and standard deviation?

The mean is calculated by adding all the values in a data set and then dividing by the total number of values. The formula for standard deviation involves finding the difference between each data point and the mean, squaring these differences, adding them all together, dividing by the total number of values, and then taking the square root of the result.

What do high and low values of mean and standard deviation indicate?

A high mean and standard deviation indicate that the data points are spread out over a larger range of values. This could suggest a larger variability in the data or the presence of outliers. A low mean and standard deviation indicate that the data points are closer to the mean, suggesting a more consistent or uniform dataset.

What is the relationship between mean and standard deviation?

The mean and standard deviation are both measures of central tendency, but they serve different purposes. The mean is a single value that represents the center of the data, while the standard deviation measures how far the data points are from the mean. A higher standard deviation indicates a larger spread of data points around the mean, while a lower standard deviation indicates a smaller spread.

How can mean and standard deviation be used to compare datasets?

Mean and standard deviation can be used to compare datasets by providing a measure of central tendency and variability for each dataset. By comparing the means and standard deviations of two datasets, we can determine which dataset has a higher or lower average and how spread out the data points are around the mean. This can help us identify any similarities or differences between the two datasets.

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