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Approximate The Mean And Standard Deviation For Age. 10 Pts

Approximate the mean and standard deviation for age. 10 pts?

The following data represent the number of people aged 25 to 64 years covered by health insurance (private or government) in 2003. Approximate the mean and standard deviation for age.
Age 25-34 |35-44 |45-54| 55-64
Number (millions) 28.7 |38.6 | 38.7 | 26.9
(type an integer or round to 2 decimal places as needed)

Calculate the mean and the standard deviation.?

That's the same as asking to find mean and the standard deviation of this data set:

64.5 74.5 74.5 74.5 74.5 74.5 74.5 84.5 84.5 84.5 84.5 84.5 94.5 94.5

which you can easily do by entering the data points in your calculator in stat mode, or by searching for "standard deviation calculator" on line and entering them there.

Statistics problem Approximate the mean and standard deviation of annual number of days over 100 degrees Fahrenheit. 10 pts?

How do I solve this? Thank you!

The following data represent the annual number of days over 100 degrees Fahrenheit in a city from 1905 to 2004. Approximate the mean and standard deviation of annual number of days over 100 degrees Fahrenheit. Treat the data as a population.
Number of over 100 degree F days 0-9 10-19 20-29 30-39 40-49 50-59 60-69
Number of years 30 33 19 11 4 2 1

What is standard deviation?

I was sure that you would get some very clever answers, with lots of maths in them. However, it looks as if I am first, and as a plain old doctor, I can give you a plain old answer!Standard deviation is a number which represents the spread of individual data in a large number of data - the divergence.Example. Suppose you measure the height of 1,000 men aged 35 in a community, and plot the results on a graph. You should end up with a bell-shaped curve, the few men who are tiny at one end, and the few giants at the other. The majority will be in the middle, the highest point of the curve.This allows you to get all sorts of wonderful (mathematical) things, but in the first instance shows you the “normal” height of a male in that community, and gives you an idea of the distribution of heights.Example. I am on a strict cardiac care regime (no surprise at my age), part of which involves taking my blood pressure and weight every morning. My systolic pressure ranges from 141 on a bad day to 117 on a good day (classic - 120). My diastolic pressure goes from 84 to 71 (classic - 80) The systolic pressure is subject to all sorts of things, including stress, temperature, and general health, or what I have just been doing. Every fifteen days, I process my figures (Excel sheet) to get the average, and also the SD for each value. The average for the systolic pressure is 121, but the SD is around 10, showing me that the pressure is very variable … so, I am prompted to ask why. The diastolic pressure averages 74, with an SD of about 3 … lowish, but fairly stable: the “skirts” of the graph for that would be quite close together, giving a high spike rather than a bell shape.. My weight averages 69 Kg, and the SD is 0.4, showing that I am neither losing nor gaining much weight at all.I hope that these two simple examples are helpful to you. Development of all this can yield the degree of confidence in other medical data, and point to areas which need investigation.

Statistics Problem Estimating Mean and Standard Deviation?

You have not got the individual specific gravities so take an
mid-point value of the intervals so
Specific gravity(x)----------------requency(f)
5.25----------------------------------...
5.45----------------------------------...
5.65----------------------------------...
5.85----------------------------------...
There are 30 values in all (9+6+11+4)
A) The estimate of the mean, M , is
(1/30)Σxf = (1/30){5.25X9+5.45X6+5.65X11+5.85X4)
which I leave you to calculate.

B) This estimate s = √[(1/29)Σf(x-M)^2]

Question about mean and standard deviation? 10 pts?

To find the percentage of students that were kicked out, you need to get the z-score corresponding to a 2.00 gpa. To get this, use the equation z = (X - μ)/σ where σ is the standard deviation and μ is the mean.

z = (2 - 2.87)/.34
z = -2.559

Now, to find the percentage that lie below that z-score, use your standard normal distribution table or calculator and find the percentage that corresponds to z = -2.559. The answer is 5.25%. For a TI-83 and above, type:
2nd
vars
2
-100
,
-2.559
)

To find the middle 90%, you need to take off 5% on each side, so you need to find the z-scores for .05 and .95. Do this using the appropriate table for standard normal distributions in your textbook, or put it into a calculator. The z-score for .05 is -1.645 and for .95 is 1.645. For a TI-83 and above, type:
2nd
vars
3
.05 [or .95]
)

Now you need to reverse-standardize for each z-score using the equation z = (X - μ)/σ where σ is the standard deviation and μ is the mean.

-1.646 = (X - 2.87)/.34
X = 2.310

1.646 = (X - 2.87)/.34
X = 3.430

The middle 90% lie between 2.310 and 3.430.

Why we should use standard deviation over mean? What type of problems can we solve with standard deviation?

These two statistics tell about two different characteristics of a random variable, so rather than saying one should be used over another, you might instead say they serve two different purposes. The question is analogous to asking why we should use a scale over a tape measure. The mean tells you what is a "typical" value of the variable being studied, or what would be the best guess for a random observation of that variable. The standard deviation tells you how large is a "typical" deviation from that mean. If you picture a histogram of your variable, the mean tells you the position on the number line upon which the distribution is centered (the center of mass... where you could balance the histogram). The standard deviation a measure of how spread out the values are around that center point. I think the best way to pull the two ideas together is to think in terms of an interval. So let's say that we know the height in inches of the male north american abominable snowman has a mean of 100 and SD of 8, and furthermore that the distribution of heights is approximately normal (bell-shaped). Those three nuggets (100, 8, normal) each tell you completely different info, but together they tell you basically everything about the distribution of the heights of male abominable snowmen. You could draw a nearly perfect sketch of this distribution using these three facts. You also know that a typical height is 100 inches, most  (68%) are between 92 and 108 inches tall, and nearly all (95%) are between 84 and 116 inches tall. Notice that the mean of 100 told you nothing about how big to make your intervals. If I had neglected to give you the SD, you would have no idea how wide to draw your curve. If you happen upon an adult male abominable snowman that's taller than 124 inches (10'4"), you should consider that to be a truly rare event.

What is the sum of average and standard deviation?

Here is a detailed example for your doubt.,Consider a set X of numbers 5,10,15,20,25                             Step 1 :         Mean = Sum of X values / N(Number of values)                  = (5+10+15+20+25) / 5                  = 75 / 5                  = 15                                                  Step 2 :          To find the variance,             Subtract the mean from each of the values,             5-15 = -10             10-15 = -5             15-15 = 0             20-15 = 5             25-15 = 10 Now square all the answers you have got from subtraction.             (-10)2 = 100             (-5)2   = 25             (0)2     = 0             (5)2    = 25             (10)2 = 100                          Add all the Squared numbers,             100 + 25 + 0 + 25 + 100 = 250                          Divide the sum of squares by (n-1)             250 / (5-1) = 250 / 4 = 62.5                          Hence Variance =  62.5 Source:Standard Deviation Calculator

What does it mean when the standard deviation is higher than the mean? What does that tell you about the data?

With respect, I disagree with Robert.Although the standard deviation is used as a unit of measurement on the normal distribution, that is not its sole function. It is more generally a measure of spread. The mean is a measure of location.The ratio of the standard deviation to the mean is called the coefficient of variation. This can be helpful in distinguishing two data sets that are given in different units. For example if we are comparing economic data from two different countries, the data might be quoted in different currencies. The coefficient of variation has no units and so a direct comparison is possible in circumstances where the mean and standard deviation cannot be directly compared.A similar ratio is the index of dispersion, which is the square of the standard deviation divided by the mean. In biology, for example, we might look at the locations of trees within a wood. An index of dispersion equal to 1 would indicate a dispersal consistent with a Poisson process, i.e. uniformly at random. A value greater than 1 would indicate greater variability than would be expected with a Poisson process; a value less than 1 would indicate less variation.In image processing, the signal-to-noise ratio can be calculated as the mean divided by the standard deviation (i.e. the reciprocal of the coefficient of variation). It can be used to determine the reliability of apparent image details.So there is a variety of circumstances under which we might compare the size of the mean and the standard deviation and derive useful results.

What does standard deviation (SD) tell about data? If the mean is 20 and the SD is 2.5, then what is meaning of 2.5 with respect to the mean (20)?

Well! You want to know , what is the meaning of SD with respect to the mean.SD is calculated, as it helps us to know how spread out the numbers are in the data. SD will be higher if the data point is very far from the mean. Very far means, the data will be more spread out. Here data point means a single fact of the data, which is normally high lighted in the data.So, you can understand that SD is very useful measure.. Here i'm giving an example. You can comprehend well then…Like, we have a data of score earned by a small group of players.. 600, 470, 170, 430, 300.When we calculate its mean= (600+470+170+430+300)/5 = 394 .Did you notice? There is a player with individual score 600. Almost double of mean.. and one player, who scored 170, almost half of the mean..If we see like that, each score differs from the mean..600–394= 206470-394= 76170-394 = mod -224430–394 = 36300–394 = mod - 94Now, you calculate SD, with above figures( i'm not showing the method here) SD= 147.32Which is quite a big measure with respect to the mean of the data. That means, the above differences are more spread out . And using this SD , we have a standard way of knowing what is normal score, & what is huge score & what is very small score…So, as you asked in the question 2.5 is SD of the data, which says that the data has moved apart by 2.5 unit from its mean 20. And this dispersion is not very high…One more thing… Statistics is all about approximate study.. :)So need not get into logic much.. as compared to all other branches of mathematics…

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