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Mean of the set = (23+24+36+38+53+56)/6 = 38
Hence if we drop the number which has the greatest distance from the mean, we will see Most decrease in std dev.
And if we drop the most near central number (not the number equal to mean), we can obtain Most increase in std dev.

Answer: Most decrease = 56, Most increase = 34
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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, Blend Green has a nutrient ratio of N:P:K = 3:2:1.
Total weight/unit = 6g

The limiting nutrient will be the one with max proportion in the blend.
N: 120/3 = 40 units
P: 90/2 = 45 units
K: 60/1 = 60 units

Therefore, the limiting nutrient is N.

Converting 40 units to g = 40*6 = 240g

Blend Rich: N:P:K =2:1:2
Total weight/unit = 5g

Limiting nutrient:
N: 120/2 = 60 units
P: 90/1 = 90 units
K: 60/2 = 30 units

Its P. Converting 30 units to g = 30*5 = 150g

Max Blend rich = 150g
Max Blend green = 240g
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Most decrease = 56
Most increase = 38

1. Identify the structure of the data and meaning of standard deviation i.e. it measures spread around the mean, so removing a value affects SD based on how far that value is from the centre.
2. Determine which value is farthest from the centre. Given Mean of the data is 38 therefore 56 is the farthest from the mean. Removing the most extreme value reduces overall spread the most. Hence greatest decrease in standard deviation.
3. Determine which value is closest to the centre. Discarding 38 which is the mean of the data, makes the remaining data more spread out relative to the new mean. This produces the greatest increase in standard deviation.
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One way to simplify the calculation would be to subtract the smallest number from all the numbers (23 in this case)

24 38 53 23 56 34 becomes
1 15 30 0 33 11

Now the mean is 15.

Most decrease will happen when you remove the farthest number from the mean i.e. 33 -> 56
Most increase will happen when you remove the closest number to the mean i.e. 15 -> 38
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Standard deviation measures how spread out values are from the mean.

Step 1: Find the Mean
Data: {24, 38, 53, 23, 56, 34}

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

Step 2: Understand the Core Logic
The key insight:
• Values FAR from the mean → increase standard deviation

• Values CLOSE to the mean → decrease standard deviation



Therefore:

• Remove a value FAR from mean → SD decreases

• Remove a value CLOSE to mean → SD increases


Step 3: Calculate Distance from Mean ([b]38)[/b]
| Value | Distance from Mean |
|-------|-------------------|
| 23 | |23 - 38| = 15 |
| 24 | |24 - 38| = 14 |
| 34 | |34 - 38| = 4 |
| 38 | |38 - 38| = 0CLOSEST |
| 53 | |53 - 38| = 15 |
| 56 | |56 - 38| = 18FARTHEST |

Step 4: Apply the Logic
Most Decrease: Remove the value FARTHEST from mean
56 is farthest (distance = 18)
→ Removing 56 causes the greatest decrease in SD

Most Increase: Remove the value CLOSEST to mean
38 is closest (distance = 0, it IS the mean!)
→ Removing 38 causes the greatest increase in SD

Answer: Most decrease = 56, Most increase = 38
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