Arithmetic mean and standard deviation

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The discussion focuses on calculating the arithmetic mean and standard deviation of days lost to sickness using a frequency distribution. The user calculates the mean as 7.85 and the variance as 16.1775, leading to a standard deviation of approximately 4.02. There is clarification that the term "taking the medians" should be referred to as "using the mid-points of the classes," which is essential for weighted means. The user expresses frustration with their university's terminology and seeks confirmation of their calculations. The forum serves as a valuable resource for support before their final exam.
LDC1972
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Homework Statement



Showing all your working, calculate the arithmetic mean and standard deviation of the number of days lost.

Table shows man days lost to sickness..

Days lost: 1-3 4-6 7-9 10-12 13-15
Frequency 8 7 10 9 6

Homework Equations


n
They've given us this Ma = Ʃ = Xi / N
I=1

The Attempt at a Solution



The textbook they've supplied is really poor. They go into detail on all values centred around a single array. Obviously mine is a double...

So using Google and previous experience I've done this:

Taken medians of Days lost=

1-3 4-6 7-9 10-12 3-15
2 5 8 11 14 (medians)

median of frequency of days lost = 8 (central number as odd amount of numbers)

Multiplied median of days lost by frequency - (2x8) + (5x7) + (8x10) + 11x9) + (14x6) = 314

Summed frequency values 8 + 7 + 10 + 9 + 6 = 40 (Total frequency)

Divided 314/40 = 7.85 Arithmetic mean

Calculated variance
= median of days lost - Arithmetic mean

i.e 2 - 7.85 = 5.85 etc

Squared all results removing negative values to positive I.e -5.85^2 = 34.2225, did same for rest.

Multiplied squares by frequency i.e 34.2225 x 8 = 273.78 etc through the 5 values

Added all multiples 647.1

Divided multiples by 40 (the frequency) 64701/40 = 16.1775

Therefore σ2 = 16.1775 (The variance)

Calculated standard deviation by √16.1775 = 4.02212603

Basically if someone would be o kind as to confirm my approach and answers that'd be great But I don't expect anyone to spend hours on this!

Thanks for help if you can.

Lloyd
 
Last edited:
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What you refer to as "taking the medians" is referred to in this setting as "using the mid-points of the classes" (or other writer-dependent language). The calculations then proceed as you've done.
The calculation for finding the mean is actually a use of the idea of "weighted means" - you may have luck finding more information by searching for that term.
 
statdad said:
What you refer to as "taking the medians" is referred to in this setting as "using the mid-points of the classes" (or other writer-dependent language). The calculations then proceed as you've done.
The calculation for finding the mean is actually a use of the idea of "weighted means" - you may have luck finding more information by searching for that term.

Thank you for the really fast reply.

I've stuck with the university's terminology. For my last assignment I didn't get a single question wrong but only got a merit, not a distinction - probably because I didn't follow their nonsense.

Others I've painfully followed their way and got distinctions.

So I'm sticking as close as I can to their terminology but agree with you 100%.

Trouble I my tutor is only available Fridays and Mondays and I missed him today. This is the lat exam of this module before exit interview and I'm going to voice my concerns!

Having this forum for the rest of the time is invaluable.

Thank you!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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