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Let's calculate the average of the given values.
The average is 38.

For maximum increase, we can remove a value with the least distance from the mean. We have 38, which is the same as the mean, and the distance is 0, so this will give the maximum increase.

For maximum decrease, we remove a value with the greatest distance from the mean. 56 is 18 units away from 38, which is the maximum(23 is just 15 units away); this will give the maximum decrease.

So 56 for maximum decrease and 38 for maximum increase.
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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}

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.
arrnaging in noncreasing order
23 24 34 38 53 56
mean is 38

greatest decrease is when we remove farthest element from mean = 56

so greatest increase would be if we remove close to mean = 38

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.
The six bottle reading volumes in milli litres are {23, 24, 34, 38, 53, 56}.

The sum = 228

Mean = 228/6 = 38

Standard Deviation (S.D) is the measure of how spread the data is, with respect to mean.

Case 1:

Now we need remove a data, which makes the SD increase , SD decrease also.

So, if we remove a value, which is closest or equal to mean. The spread increases.

Then, removing 38, which is the mean, will result in the mean =38.

But, the values around the mean is diminished, there by pushing the spread (S.D).

S.D INCREASE most happens at value 38.

Case 2:

When we remove the data which is at the farthest distance from the mean (38) = 56.

This pushes the values towards the mean. New mean = 34.4.

The values are pulled towards the mean, with the left side having more spread than the right side. The skewness is more towards the left hand side.

Thus, the S.D DECREASES most at 56.

Most S.D increase = 38

Most S.D decrease = 56



For general understanding :

Imagine a normal bell shaped curve, or a tent in bell shape, which has a long pole at the centre (mean).

Values closer or equal to mean:

When you add more data at the mean position or closer to the mean, the clustering of data happens at that point. This increases the height of the curve, thereby pulling the outlier data towards the mean. So, the spread decreases.

When you remove values which are equal to mean or closer to mean, assume u remove all structures which is holding the curve as bell shaped near the mean. This will result in data falling, and the spread moving across in both directions, thus spread increases.

Values away from mean:

When you add more data away from mean, your stretching the curve beyond its actual shape, so the spread increases.

When you remove data away from mean, your releasing the tension held by the values, so that it shrinks to it original shape. So, spread decreases.
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We have the set {24,38,53,23,56,34} we need to find which number if removed will result in most decrease and increase in standard deviation. Ideally if we arrange the numbers in increasing order of set , it will become {23 ,24,34,38,53,56} . Now seeing this set if we remove 56 then the numbers will become more compact and standard deviation will be less since the mean of this set is 38 and distance between 38 and 56 is highest. And as we move close to mean , standard deviation will start decreasing . If we remove 38 , the mean still remains same and standard deviation will most increase because variance will be larger.

So answer is Remove 56 to get most decrease in SD. Remove 38 to get most increase in SD.
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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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Mean of the series = 24+38+53+23+56+34/6 = 228/6 = 38

Most decrease is when we remove the value farthest from the mean i.e. 56
Most increase is when we remove the value closer to the mean i.e. 38

Answer=> Most decrease => 56, most increase => 38
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{24, 38, 53, 23, 56, 34}
{23, 24, 34, 38, 53, 56}

Standard deviation is a quantity expressing by how much the members of a group differ from the mean value for the group.

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

Most decrease
Standard deviation is most reduced by removing the value farthest from the mean, since that value contributes most to the spread.
56-38=18 > 38-23=15
So, remove 56.

Most increase
Removing a value close to the mean makes the remaining data more spread out.
The mean, 38, is one of the values.
So, remove 38.

Most decrease=56
Most increase=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.

Results: (23,24,34,38,53,56)
Mean= (23+24+34+38+53+56)/6= 228/6= 38

We know that removing a value far from mean, descreases SD the most.
And removing a value close to mean, increases SD the most.

Distance of all points from mean,
38-23= 15
38-24= 14
38-34= 4
38-38= 0 (nearest to mean)
38-53= 15
38-56= 18 (farthest from mean)

Most decrease in SD= 56
Most increase in SD= 38
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arithmetic mean of the values is 38

Standard deviation measures how much the individual values in a dataset spread out from the average. A low standard deviation means most numbers are close to the mean and a high one means they are more scattered.

Removing the value that is most distant from the average shrinks the spread the most.
The largest deviation is 56 (56-38=18), so removing it produces the greatest decrease in standard deviation.

If you remove a value close to the mean, the leftover data becomes more dispersed.
Discarding 38 removes the central value and leaves the remaining data more dispersed, producing the greatest increase in standard deviation.

Most decrease=56 and Most increase=38
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average of the six measurements = 38

Standard deviation quantifies how widely individual values deviate from the average.

The value furthest from the mean is the biggest contributor to spread, so discarding it decreases standard deviation the most.
The furthest value can be 56 (56-38=18) or 23 (38-23=15).
The answer is 56

Deleting a number that is near the mean causes the rest of the data to be more spread apart.
We can remove the mean, 38.

Most decrease=56
Most increase=38
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Mean = (24+38+53+23+56+34)/6 = 38

Standard deviation tells you the typical distance between each data point and the mean of a dataset.

Standard deviation drops the most when you take out the number farthest from the mean.
The number farthest from 38 is 56.

Taking out a value near the average often makes the remaining set less clustered, increasing variability.
38 is not only near the average, it's the average.

Most decrease=56 and Most increase=38
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Mean of given 6 terms = 38

Removing an extreme value have greatest decrease on SD. Here it is 56.

And removing 38 (the mean) is going to have the greatest increase on SD compared to rest of the numbers.
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given six readings {23,24,34,38,53,56} of which any one value is to be discarded.
mean of the above values is 38.
since SD=square root of[ {summation of the squares of differences of each values from the mean}/total no of values]
hence discarding the value which is farthest from the mean will decrease the SD most and vice versa.
hence most decrease=56 which is 18 units (maximum) away from 38
and most increase=38 which is same as the mean.
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To solve this without heavy calculations, we can use Standard Deviation (SD).
Standard deviation will measure how far the data points are from the mean.

Sum = 23 + 24 + 34 + 38 + 53 + 56 = 228
Mean = 228/6 = 38

Most decrease in standard deviation will reduce SD the most, remove a value farthest from the mean (an extreme).
Distances from mean: 23 → 56 → farthest
Removing 56 shrinks the spread the most. Most increase in standard deviation To increase SD, remove a value near the center, leaving extremes farther apart.

Closest to mean: 38
Removing 38 increases spread the most.

Final Answer Most decrease: 56
Most increase: 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}

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.

SD indicates the position or distance of the elemets from the mean

Mean = (24+38+53+23+56+34)/6 = 38

Placement of elements

23 --- 24 -------- 34 -----38 -------------------53 -----56

Hence, if we remove the element placed at 38, the SD will increase and if we remove the element placed at 56, the SD will decrease.

Most decrease : 56
Most increase : 38
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Most decreased-56 by removing the farthest value from the mean shrinks and spread the most.

Most increased-38
X=(24,38,53,23,56,34)
mean=38
std deviation=14.04
23=13.38
24=13.7
34=15.54(increase)
38=15.7(increase)
53=13.38
56=12.22(decrease)
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Standard deviation is (x - Xmean)^2/n

So we know the mean of 6 would be 38

So since the number is in square, so there will be no difference is numbers are less than 38 or not

So if we see number 56 if we remove that, since the difference between 56 and 38 is highest like 18
if we remove this then sum would decrease and also the standard deviation

So for largest decrease we can remove number 56

and for increase if we remove 38 and mean is 38 so we are removing this will not impact sum and it will increase the standard deviation so for increase we remove 38

Hence answer 56 for decrease and 38 for increase.
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We are given 6 numbers which are {23,24,34,38,53,56}
We need to find the number when eliminated would give greatest decrease in SD from the original SD and also for the greatest increase in SD from the original SD

Mean = 23+24+34+38+53+56 / 6 = 38

Now, we see that 56 is the farthest from the mean. Upon eliminating it we would get the greatest decrease from the original SD and a decrease in the value spread.

Now since we know that mean is 38, removing the number 38 which is central point would cause the most increase in the original SD

Most decrease : 56
Most increase : 38
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