how many players are in the range of one standard deviation from the mean.

Question

what is one standard deviation

Answer 1

Ok imagine it is the final run. The trainer has the choice for the last position in the team lets say between Carl and Jim.
He has all the data of them.
The median is equal: 11,2 seconds/100 meters.
But Jim sometimes is very fast and on some days he is very slow. So he has a high standard deviation.
Carl is like a machine: 11,2 11,1 11,3 11,2 11,2
He is a constant om two feet with a low standard deviation.

Now if trainer takes risk he chooses Jim.
For safety Carl is the right man.

Answer 2

Frogged , i think that your explanation for v(ariance) is missing the division at the end of the expression by the number of all players?

v=sum (value-mean) **2 of all players / number of players

Answer 3

Harsh Mandwariya , to get the standard deviation, you need to calculate the mean, then the variance, and finally the standard deviation.

▪︎take the mean, and subtract the std dev to get the lower bound
▪︎take the mean again and add the std dev to get the upper bound
▪︎the range between these both values is that what is mentioned as "one standard deviation". (mean ± std dev)
▪︎finally you have to check how many players are in this range.

happy coding!

Answer 4

get mean
v=sum (value-mean) **2 of all players
standard deviation = root(v)

Answer 5

yeah... right

Answer 6

I think it was given as an example in the lesson that one standard deviation means the number is within the range from (mean - std) up to (mean + std)

Answer 7

Frogged , what a nice example . I understood the term completely. But I just wanted to know that is there something like it two standard deviation ir 3 standard deviation exists just like 1 standard deviation

Answer 8

I know these things . I just wanted to know what do one standard deviation refers to

Answer 9

Lothar , Thank you so much

Answer 10

players = [180, 172, 178, 185, 190, 195, 192, 200, 210, 190]
mean=(sum(players)/len(players))
l=[]
for i in players:
l.append((i-mean)**2)
stdev=pow(sum(l)/len(players),0.5)
a=mean+stdev
b=mean-stdev
c=0
for i in players:
if i>b and i<a:
c=c+1
print(c)

Answer 11

players = [180, 172, 178, 185, 190, 195, 192, 200, 210, 190]
dev stdev(players):
n = len(players)
mean = sum(players)/ n
deviation = [(x-mean)** 2 for x in players]
variance = sum(deviation)/n
stdev = variance ** (1/2)
i = mean - stdev
j = mean + stdev
for s in players:
if s>i and s< j:
print(s)

print(stdev(players))

i did this but its not real answer should i calculate the sum of players?