Thank you so much for the kind words! We're not sure why, either -- it's a relatively new channel, so hopefully the news will spread eventually. :) Have fun studying, and thank you again!
Can we solve question 7 algebraically? Since 65 is the mean, Can't we say that 65 + 10/17(SD) = 85 (Since 65 + SD would give the weight at 84th percentile) ? Solving this, we would get SD as 34. Since we need the 60th percentile, wouldn't the value be equal to 65 + 5/17 (SD) = 65 + 10 =75. May I know if there's anything wrong in this approach? Thank you!
This depends a little on what you mean by solve the Standard Deviation. If we know the standard deviation of a set of numbers, and we know that every element of that set has been multiplied by the same number (let's call it x) to create a new set, then the standard of the new set will be the standard deviation of the old set multiplied by x. Multiplying every element of a set by a number (let's keep using x) changes the standard deviation of that set by a factor of x. However, if we know the standard deviation of a set of numbers, and we know that we added the same number (let's call it y) to every element of that set to create a new set, then the standard of the new set will be the same as the standard deviation of the old set. Adding or subtracting a number to every element of a set does not change the standard deviation of that set because it doesn't change how spread out the numbers are. I hope that helps!
Dont you think that SD of T would be lower because whatever the unknown numbers would be in list T their Mean would be higher therefore the SD would be lower for list T compared to list S as their mean would be relatively lower therefore by hypothetical comparison in formula of SD, higher mean would result in lower SD. Would like your input. Your videos have helped me a lot to prepare.
In a very rough, basic, unscientific, hand-wavy way, you can think of standard deviation as a measure of how spread out the numbers are in a given set. The value of the mean doesn't matter all that much. While we'd use the mean in the formula for the standard deviation if we really wanted to calculate the standard deviation for a set, whether the mean is a large number or a small number doesn't actually affect the standard deviation. What affects the standard deviation is how spread out the numbers are or, on average, how far are the numbers from the mean. In this question, we know all the numbers in List S, so we could figure out the standard deviation of List S if we wanted to. For List T, we know the greatest and least number in the set, so we know the range of the set is 12. This is the same as the range of the numbers in List S, but we know nothing about how spread out the numbers are in List T. We don't even know how many numbers are in List T, but let's assume it has the same number of elements as List S. If List T was: 131, 137, 137, 137, 137, 137, 143 then the standard deviation of List T would be smaller than the standard deviation of List S as the numbers in List T would be closer to the mean, on average, than the numbers in List S. However, if List T was: 131, 131, 131, 137, 143, 143, 143 then the standard deviation of List T would greater than the standard deviation of List T as the numbers in List T would be further away from the mean, on average, than the numbers in List S. Since we can't tell whether the standard deviation of List T is greater than or less than the standard deviation of List S, the answer to this question is (D). I hope that helps!
Sd= racine( sum((Xi-mean)**2)/N) If you add the same nu,ber to the set, the sd dosnt change, because you dont changebthe way the databis spread ou If you multiply the sd changes
In The normal distribution graph,the area under the curve between two points represents the probability of observing values within that range, in relation to median/ mean and sd If x is normaly distributed then 34% of data is within [mean, mean+sd] and 14% it s between [mean+sd, mean+2sd] and only 2% it bigger than ,ean+2sd To get a comparaison relation between the percentiles and the normal distribution graph
The last answer does not make logical sense. I know it's a rough calculation, but I don't understand how it was calculated. It's just a rough drawing, how can we know anything for sure
Because had he made it at exactly 75kg, the shapes made would not be equal. This is a rough calculation as you could see. If he put it at 75, the shape formed by 65-75 would be greater than that of 75-85 and they need to be equal. Trying to split a disproportionate object is tricky but just knowing you would have to be slightly left would net you the correct answer.
![Quartiles, Percentiles, Quintiles, Interquartile Range, Box Plots [worked examples! Oh wow]](http://i.ytimg.com/vi/fhFDjxcOrb8/mqdefault.jpg)
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dunno why it has so less views and counts but one of the best channel for gre I have seen
Thank you so much for the kind words! We're not sure why, either -- it's a relatively new channel, so hopefully the news will spread eventually. :)
Have fun studying, and thank you again!
@@GRENinjaTutoring Aaa, Your Welcome :)
Same shirt, same grind, heck ya :000
Haha! Same luscious accent, too. 😆
@@GRENinjaTutoring Yummy Irish man 🔥😍🔥😍
Q4 makes more sense (to an average person) to start with cross multiplication, took me less than a minute to calculate everything.
Can we solve question 7 algebraically? Since 65 is the mean, Can't we say that 65 + 10/17(SD) = 85 (Since 65 + SD would give the weight at 84th percentile) ? Solving this, we would get SD as 34. Since we need the 60th percentile, wouldn't the value be equal to 65 + 5/17 (SD) = 65 + 10 =75. May I know if there's anything wrong in this approach? Thank you!
With your explanation on question 2, can we solve the SD by just multiplying ?
This depends a little on what you mean by solve the Standard Deviation.
If we know the standard deviation of a set of numbers, and we know that every element of that set has been multiplied by the same number (let's call it x) to create a new set, then the standard of the new set will be the standard deviation of the old set multiplied by x. Multiplying every element of a set by a number (let's keep using x) changes the standard deviation of that set by a factor of x.
However, if we know the standard deviation of a set of numbers, and we know that we added the same number (let's call it y) to every element of that set to create a new set, then the standard of the new set will be the same as the standard deviation of the old set. Adding or subtracting a number to every element of a set does not change the standard deviation of that set because it doesn't change how spread out the numbers are.
I hope that helps!
Dont you think that SD of T would be lower because whatever the unknown numbers would be in list T their Mean would be higher therefore the SD would be lower for list T compared to list S as their mean would be relatively lower therefore by hypothetical comparison in formula of SD, higher mean would result in lower SD.
Would like your input. Your videos have helped me a lot to prepare.
In a very rough, basic, unscientific, hand-wavy way, you can think of standard deviation as a measure of how spread out the numbers are in a given set. The value of the mean doesn't matter all that much. While we'd use the mean in the formula for the standard deviation if we really wanted to calculate the standard deviation for a set, whether the mean is a large number or a small number doesn't actually affect the standard deviation. What affects the standard deviation is how spread out the numbers are or, on average, how far are the numbers from the mean.
In this question, we know all the numbers in List S, so we could figure out the standard deviation of List S if we wanted to. For List T, we know the greatest and least number in the set, so we know the range of the set is 12. This is the same as the range of the numbers in List S, but we know nothing about how spread out the numbers are in List T. We don't even know how many numbers are in List T, but let's assume it has the same number of elements as List S. If List T was:
131, 137, 137, 137, 137, 137, 143
then the standard deviation of List T would be smaller than the standard deviation of List S as the numbers in List T would be closer to the mean, on average, than the numbers in List S. However, if List T was:
131, 131, 131, 137, 143, 143, 143
then the standard deviation of List T would greater than the standard deviation of List T as the numbers in List T would be further away from the mean, on average, than the numbers in List S.
Since we can't tell whether the standard deviation of List T is greater than or less than the standard deviation of List S, the answer to this question is (D).
I hope that helps!
22:40 , @@GRENinjaTutoring can you pls tell how to complete the graphical method solution to arrive at the final answer
Sd= racine( sum((Xi-mean)**2)/N)
If you add the same nu,ber to the set, the sd dosnt change, because you dont changebthe way the databis spread ou
If you multiply the sd changes
Percentile of observation x = rank of x on ascending order/N *100 where N number of obs
Median: 50% percentile
Quartiles 25%...
Quintles:20%,....
Deciles: 10%
Using percentiles in alpt of situation, use percentioemof value x is pr means that pr*N have lesser value than x
In The normal distribution graph,the area under the curve between two points represents the probability of observing values within that range, in relation to median/ mean and sd
If x is normaly distributed then 34% of data is within [mean, mean+sd] and 14% it s between [mean+sd, mean+2sd] and only 2% it bigger than ,ean+2sd
To get a comparaison relation between the percentiles and the normal distribution graph
The last answer does not make logical sense. I know it's a rough calculation, but I don't understand how it was calculated. It's just a rough drawing, how can we know anything for sure
Sorry, I got it now.
49:07 why you put the line left to the 75kg can you please explain?
Because had he made it at exactly 75kg, the shapes made would not be equal. This is a rough calculation as you could see. If he put it at 75, the shape formed by 65-75 would be greater than that of 75-85 and they need to be equal. Trying to split a disproportionate object is tricky but just knowing you would have to be slightly left would net you the correct answer.