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A quality inspector recorded the fill volumes, in milliliters, of six bottles from a production run, with the following results:

{24, 38, 53, 23, 56, 34}

The inspector believes exactly one of the six measurements is erroneous and will discard that measurement. To evaluate the impact of the discard, he will calculate the standard deviation of the original six measurements and the standard deviation of the remaining five measurements.

Select for Most decrease the measurement which, if discarded, would produce the greatest decrease in standard deviation, and select for Most increase the measurement which, if discarded, would produce the greatest increase in standard deviation. Make only two selections, one in each column.

values
23, 24,38,53,56,34

mean of set is 228/6 ; 38

re arrange set
23,24,34,38,53,56
removing 38 will provide most increase in SD as the value along mean will increase

removing 56 will provide most decrease in SD as the value along mean will decrease

most decrease 56, most increase 38 is correct
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answer is 56 and 38

using the formula oof std deviation ((x-u)^2/sigma)^1/2 we can find the answer
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When it comes to the correlation between the standard deviation and the mean, add/removing a value closest to the mean has the least impact on the standard deviation while adding or removing the most extreme value has the greatest impact on the standard deviation
From the above we can clearly see the greatest decline will come from the removal of 56 (the most extreme value) and the most increase will come from the removal of a value closest to the mean which is 38
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A quality inspector recorded the fill volumes, in milliliters, of six bottles from a production run, with the following results:

{24, 38, 53, 23, 56, 34}

The inspector believes exactly one of the six measurements is erroneous and will discard that measurement. To evaluate the impact of the discard, he will calculate the standard deviation of the original six measurements and the standard deviation of the remaining five measurements.

Select for Most decrease the measurement which, if discarded, would produce the greatest decrease in standard deviation, and select for Most increase the measurement which, if discarded, would produce the greatest increase in standard deviation. Make only two selections, one in each column.
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A quality inspector recorded the fill volumes, in milliliters, of six bottles from a production run, with the following results:

{24,38,53,23,56,34}

Arranging the numbers in increasing order = {23, 24, 34, 38, 53, 56}

Average (arithmetic mean) = (23+24+34+38+53+56)/6 = 38

Distance from mean = {15, 14, 4, 0, 15, 18}

To increase the standard deviation we should remove data closest to the mean = 38 so that data set appears to be spread out.
To decrease the standard deviation we should remove data farthest from the mean = 56 so that data set appears to be shrinking in.

Most decrease56
Most increase38
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the standard deviation for all 6 measurements is 36

when the error is removed for most decrease in standard deviation, it occurs when greatest value of measurement is removed standard deviation drops from 36 to 34 at max

similarly it happens when lowest measurement value is removed, standard deviation value increases to 38 from 36.

hence the selections are 56 & 23
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A quality inspector recorded the fill volumes, in milliliters, of six bottles from a production run, with the following results:

{24, 38, 53, 23, 56, 34}

The inspector believes exactly one of the six measurements is erroneous and will discard that measurement. To evaluate the impact of the discard, he will calculate the standard deviation of the original six measurements and the standard deviation of the remaining five measurements.

Select for Most decrease the measurement which, if discarded, would produce the greatest decrease in standard deviation, and select for Most increase the measurement which, if discarded, would produce the greatest increase in standard deviation. Make only two selections, one in each column.
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Considering the definition of standard deviation, you will get the greatest decrease when you remove the number that's furthest from the mean of the set.
On the contrary, you will get the greatest increase, when you discarded the value that's closest to the mean.

mean = 228 / 6 = 38

see which of the values has the greatest "distance" from 38.
38 - 23 = 15
56 - 38 = 16
greatest decrease: 56

greatest increase: 38
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[23,24,34,38,53,56]
Average = 38
now here 56 lies fartherest to the average 38, so removing 56 will give highest decrease in SD
If 38, which is also the average is removed, the series spreads out without changing the average. this gives greatest increase in SD.

Ans 56 & 38
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A quality inspector recorded the fill volumes, in milliliters, of six bottles from a production run, with the following results:

{24, 38, 53, 23, 56, 34}

The inspector believes exactly one of the six measurements is erroneous and will discard that measurement. To evaluate the impact of the discard, he will calculate the standard deviation of the original six measurements and the standard deviation of the remaining five measurements.

Select for Most decrease the measurement which, if discarded, would produce the greatest decrease in standard deviation, and select for Most increase the measurement which, if discarded, would produce the greatest increase in standard deviation. Make only two selections, one in each column.

We can eyeball this question by keeping in mind some rules about standard deviation. First we calculate the average, which is \(\frac{24+38+53+23+56+34}{6}=38\).

For Column 1, we need to discard a measurement that would result in a decrease in standard deviation. To decrease standard deviation, we need to get all values closer to the average, which means we need to throw out the measurement farthest from the average. Looking at the extremes, 24 is 12 units away and 56 is 18. Therefore, we remove 56, which is our column 1 answer.

For Column 2, to increase standard deviation, we need to throw away values that are close to the average. Luckily, the average is 38 and we have that as part of our set, so that is the answer for Column 2.
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A quality inspector recorded the fill volumes, in milliliters, of six bottles from a production run, with the following results:

{24, 38, 53, 23, 56, 34}

The inspector believes exactly one of the six measurements is erroneous and will discard that measurement. To evaluate the impact of the discard, he will calculate the standard deviation of the original six measurements and the standard deviation of the remaining five measurements.

Select for Most decrease the measurement which, if discarded, would produce the greatest decrease in standard deviation, and select for Most increase the measurement which, if discarded, would produce the greatest increase in standard deviation. Make only two selections, one in each column.
avg (23,24,34,38,53,56)=38 = m (say)

removing the item closes to the mean causes the ther terms of the set to be more spread about the mean and hence causes the maximum std deviation
so for maximum increase in std deviation remove x= 38
removing the term farthest from the mean brings the spread of element in the set closest to the mean
so remove x= 56 for maximum decease in std deviation
Ans= 56,38
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The mean is 38, so the number that causes the maximum change in the mean with its removal with cause the greatest standard deviation. If removing a number keeps the mean at the center hovering around 34-38 will cause the least change in standard deviation. Here the smallest number 23 causes the greatest decrease in SD and the largest number 56 causes the maximum increase in SD
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numbers in sorted order : {23, 24, 34, 38, 53, 56}

mean: \(\frac{(23 + 24 + 34 + 38 + 53 + 56) }{ 6} = \frac{228}{6} = 38\)

standard deviation is the average of distance from mean

For it to be most decrease, we need to remove that one which is farthest from mean:
in this case 56 is the farthest since 56 - 38 = 18
where w.r.t 23 it is 38 - 23 = 15

For it to be most increase, it has to be closest to the mean,
in this case the number 38 is 0 distance from mean 38

most decrease: 56
most increase: 38

ans: 56, 38 or F, D
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Mean = 23 + 24 + 34 + 38 + 53 + 56 / 6 = 38

Distance of the values from Mean
|23 - 38| = 15
|24 - 38| = 14
|34 - 38| = 4
|38 - 38| = 0
|53 - 38| = 15
|56 - 38| = 18

Since, 56 is farthest from the mean in absolute terms, it would decrease the SD of the series the most.
38 is the mean itself, therefore, removing 38 would increase the SD the most since the values will be farther from the mean now.

Hence, Most decrease = 56
Most increase = 38
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Finding the mean of all observations

( 24+38+53+23+56+34)/6 = 228/6 = 38
Most increase in standard deviation happens when the observation closest to mean is discarded. This is 38
Most decrease in standard deviation happens when the observation farthest to mean is discarded. This is 56
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A quality inspector recorded the fill volumes, in milliliters, of six bottles from a production run, with the following results:

{24, 38, 53, 23, 56, 34}

The inspector believes exactly one of the six measurements is erroneous and will discard that measurement. To evaluate the impact of the discard, he will calculate the standard deviation of the original six measurements and the standard deviation of the remaining five measurements.

Select for Most decrease the measurement which, if discarded, would produce the greatest decrease in standard deviation, and select for Most increase the measurement which, if discarded, would produce the greatest increase in standard deviation. Make only two selections, one in each column.
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By looking at formula of standard deviation the removal of farthest point from average of set will result in most decrease and removal of nearest point from average of set will increase the standard deviation most. Hence answer is 56 & 34 respectively.
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A quality inspector recorded the fill volumes, in milliliters, of six bottles from a production run, with the following results:

{24, 38, 53, 23, 56, 34}

The inspector believes exactly one of the six measurements is erroneous and will discard that measurement. To evaluate the impact of the discard, he will calculate the standard deviation of the original six measurements and the standard deviation of the remaining five measurements.

Select for Most decrease the measurement which, if discarded, would produce the greatest decrease in standard deviation, and select for Most increase the measurement which, if discarded, would produce the greatest increase in standard deviation. Make only two selections, one in each column.
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Arranging the numbers in ascending order, we have 23,24,34,38,53,56.

The average comes out to be 34.

Now for increasing the standard deviation by the maximum, we have to reduce a value closes to the mean. In our data set, we can remove the value 34 which is also the mean. This would have the maximum impact as a value as close to the mean as possible is being removed.

On the other hand, to reduce the standard deviation, we can look at the extreme values with respect to 34. We can see that 56 is farther from 34 than 23 is. Therefore, if 56 is removed, it would have the highest impact in terms of reduction in standard deviation.

Therefore, Most decrease = 56 and Most increase = 34
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num (23,24,34,38,53,56)
mean= 38
SD = (-15,-14,-4,0,15,18)
if we want most decrease then we need to remove num that is farthest from the mean. here that is 56

if we want most increase then we need to remove num that is closest to mean. here that is is mean itself 38.
because the other num will have more distance from mean.
but if we remove any other than 38,56, then the distance will not be that much spread across the num.
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A quality inspector recorded the fill volumes, in milliliters, of six bottles from a production run, with the following results:

{24, 38, 53, 23, 56, 34}

The inspector believes exactly one of the six measurements is erroneous and will discard that measurement. To evaluate the impact of the discard, he will calculate the standard deviation of the original six measurements and the standard deviation of the remaining five measurements.

Select for Most decrease the measurement which, if discarded, would produce the greatest decrease in standard deviation, and select for Most increase the measurement which, if discarded, would produce the greatest increase in standard deviation. Make only two selections, one in each column.
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Given: {23, 24, 34, 38, 53, 56} -> Mean = 228/6 = 38
SD -> spread from mean. Here we need to remove 1 number hence spread in SD both numerator and denominator will decrease.
min spread decrease -> when pick closest to mean -> 38 -> hence removing 38 will only have effect in decreasing denominator -> increase SD -> max SD.
max spread increase -> when pick farthest to mean -> 56 -> hence removing 56 will decrease both numerator and denominator but numerator delta = 56-38 = 18 -> squaring -> 324 which is much greater decrease than denominator 6->5.
hence 1st selection 56 and 2nd selection 38.
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