Answer:
Coefficient of skewness = 0.5785
Population standard deviation = 88.154
Step-by-step explanation:
Given the data:
Month Jan Feb Mar Apr May Jun Jul
Unit Sales 314 285 158 482 284 310 281
Reordered data : 158, 281, 284, 285, 310, 314, 482
The population mean of the data :
Mean, μ = Σx / n = 2114 / 7 = 302
The median :
1/2(n+1)th term
n = 7
1/2(8)th term
Median = 4th term = 285
The population standard deviation, s :
s = √(Σ(x - μ)²/n)
s = √[(158-302)^2 + (281-302)^2 + (284-302)^2 + (285-302)^2 + (310-302)^2 + (314-302)^2 + (482-302)^2] / 7
s= √(54398 / 7)
s = √7771.1428
s = 88.154
The Pearson Coefficient of skewness :
[3(μ - median)] / s
3(302 - 285) / 88.154
3(17) / 88.154
51 / 88.154
= 0.5785
The answer is 2,134 because you do 12% times 9,700 + 970
At the point when Hannah takes her first sweet from the sack, there is a 6/n chance it is orange.
This is because that there are 6 orange desserts and n desserts altogether.
When Hannah takes out her second sweet, there is a 5/(n-1) chance that it is orange.
This is because there are just 5 orange desserts let alone for an aggregate of n-1 desserts.
The possibility of getting two orange desserts in succession is the main likelihood increased by the second one: 6/n x 5/n–1
The question lets us know that the shot of Hannah getting two orange desserts is 1/3.
So: 6/n x 5/n–1 = 1/3
Now, rearrange this problem.
(6x5)/n(n-1) = 1/3
This gets to be:
30/(n² – n) = 1/3
Times by 3 on both sides:
90/(n² – n) = 1
What's more, doing likewise with (n² – n):
So (n² – n) = 90
Our answer is: n² – n – 90 = 0
