The answer is 1.715 hope I helped
Answer:
Step-by-step explanation:
From the information given,
Mark: 6,7,8,9,10
frequency:5,4,7,10,4
a) Range = highest mark - lowest mark
Range = 10 - 6 = 4
b) The number of students in the group is the sum of the frequency. Therefore,
Number of students = 5 + 4 + 7 + 10 + 4 = 30 students
c) Mean mark = (mark × frequency)/total frequency
[(6 × 5) + (7 × 4) + (8 × 7) + (9 × 10) + 10 × 4)]/ 30
Mean mark = (30 + 28 + 56 + 90 + 40)/30 = 244/30
Mean mark = 8.1
Answer:
![E(X) =1 *\frac{1}{5} +2 *\frac{2}{5} +7*\frac{2}{5}= 3.8](https://tex.z-dn.net/?f=%20E%28X%29%20%3D1%20%2A%5Cfrac%7B1%7D%7B5%7D%20%2B2%20%2A%5Cfrac%7B2%7D%7B5%7D%20%2B7%2A%5Cfrac%7B2%7D%7B5%7D%3D%203.8)
Now we can find the second moment with this formula:
![E(X^2) = \sum_{i=1}^n X^2_i P(X_i)](https://tex.z-dn.net/?f=%20E%28X%5E2%29%20%3D%20%5Csum_%7Bi%3D1%7D%5En%20X%5E2_i%20P%28X_i%29)
And replacing we got:
![E(X^2) =1^2 *\frac{1}{5} +2^2 *\frac{2}{5} +7^2*\frac{2}{5}= 21.4](https://tex.z-dn.net/?f=%20E%28X%5E2%29%20%3D1%5E2%20%2A%5Cfrac%7B1%7D%7B5%7D%20%2B2%5E2%20%2A%5Cfrac%7B2%7D%7B5%7D%20%2B7%5E2%2A%5Cfrac%7B2%7D%7B5%7D%3D%2021.4)
The variance would be given by:
![Var(X) =E(X^2) -[E(X)]^2 = 21.4 -[3.8]^2 = 6.96](https://tex.z-dn.net/?f=%20Var%28X%29%20%3DE%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%2021.4%20-%5B3.8%5D%5E2%20%3D%206.96)
And the deviation would be:
![Sd(X) =\sqrt{6.96}= 2.638](https://tex.z-dn.net/?f=%20Sd%28X%29%20%3D%5Csqrt%7B6.96%7D%3D%202.638)
Step-by-step explanation:
For this case we have the following distribution given:
X 1 2 7
P(X) 1/5 2/5 2/5
We need to begin finding the mean with this formula:
![E(X) = \sum_{i=1}^n X_i P(X_i)](https://tex.z-dn.net/?f=%20E%28X%29%20%3D%20%5Csum_%7Bi%3D1%7D%5En%20X_i%20P%28X_i%29)
And replacing we got:
![E(X) =1 *\frac{1}{5} +2 *\frac{2}{5} +7*\frac{2}{5}= 3.8](https://tex.z-dn.net/?f=%20E%28X%29%20%3D1%20%2A%5Cfrac%7B1%7D%7B5%7D%20%2B2%20%2A%5Cfrac%7B2%7D%7B5%7D%20%2B7%2A%5Cfrac%7B2%7D%7B5%7D%3D%203.8)
Now we can find the second moment with this formula:
![E(X^2) = \sum_{i=1}^n X^2_i P(X_i)](https://tex.z-dn.net/?f=%20E%28X%5E2%29%20%3D%20%5Csum_%7Bi%3D1%7D%5En%20X%5E2_i%20P%28X_i%29)
And replacing we got:
![E(X^2) =1^2 *\frac{1}{5} +2^2 *\frac{2}{5} +7^2*\frac{2}{5}= 21.4](https://tex.z-dn.net/?f=%20E%28X%5E2%29%20%3D1%5E2%20%2A%5Cfrac%7B1%7D%7B5%7D%20%2B2%5E2%20%2A%5Cfrac%7B2%7D%7B5%7D%20%2B7%5E2%2A%5Cfrac%7B2%7D%7B5%7D%3D%2021.4)
The variance would be given by:
![Var(X) =E(X^2) -[E(X)]^2 = 21.4 -[3.8]^2 = 6.96](https://tex.z-dn.net/?f=%20Var%28X%29%20%3DE%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%2021.4%20-%5B3.8%5D%5E2%20%3D%206.96)
And the deviation would be:
![Sd(X) =\sqrt{6.96}= 2.638](https://tex.z-dn.net/?f=%20Sd%28X%29%20%3D%5Csqrt%7B6.96%7D%3D%202.638)
Answer:
The number of Jack's stamps is 33 and Dylan's stamps is 13
Step-by-step explanation:
Let
x -----> the number of Jack's stamps
y -----> the number of Dylan's stamps
we know that
x+y=46 ----> equation A
x=y+20 -----> equation B
Solve by substitution
substitute equation B in equation A and solve for y
(y+20)+y=46
2y=46-20
2y=26
y=13 stamps
Fond the value of x
x=y+20 -----> x=13+20=33 stamps
therefore
The number of Jack's stamps is 33 and Dylan's stamps is 13
Answer:
11,744
Step-by-step explanation: