Reflection across x axis and move 1 unit to the right
Answer:
a. E(x) = 3.730
b. c = 3.8475
c. 0.4308
Step-by-step explanation:
a.
Given
0 x < 3
F(x) = (x-3)/1.13, 3 < x < 4.13
1 x > 4.13
Calculating E(x)
First, we'll calculate the pdf, f(x).
f(x) is the derivative of F(x)
So, if F(x) = (x-3)/1.13
f(x) = F'(x) = 1/1.13, 3 < x < 4.13
E(x) is the integral of xf(x)
xf(x) = x * 1/1.3 = x/1.3
Integrating x/1.3
E(x) = x²/(2*1.13)
E(x) = x²/2.26 , 3 < x < 4.13
E(x) = (4.13²-3²)/2.16
E(x) = 3.730046296296296
E(x) = 3.730 (approximated)
b.
What is the value c such that P(X < c) = 0.75
First, we'll solve F(c)
F(c) = P(x<c)
F(c) = (c-3)/1.13= 0.75
c - 3 = 1.13 * 0.75
c - 3 = 0.8475
c = 3 + 0.8475
c = 3.8475
c.
What is the probability that X falls within 0.28 minutes of its mean?
Here we'll solve for
P(3.73 - 0.28 < X < 3.73 + 0.28)
= F(3.73 + 0.28) - F(3.73 + 0.28)
= 2*0.28/1.3 = 0.430769
= 0.4308 -- Approximated
To find the area of this just add the units I believe which is 21
Hope this helps good luck. Not sure if this is 100% right my last year teacher was a sucky one
Answer:
Step-by-step explanation:
The given data is 11, 77, 35, 5, 76, 34, 89, 92, 22
The total number of items in the data is 9. The width of each class interval would be gotten by dividing the difference between the minimum and maximum items in the data by the number of class intervals. From the information given,
Minimum item = 5
Maximum item = 92
Difference = 92 - 5 = 7
Class width = 87/10 = 8.7
By approximation, it is closest to 10
Therefore, the appropriate class group is 0 to 10
Answer: B
Step-by-step explanation:
The p-value is greater than the significance level (0.09 > 0.05), so we "fail to reject" H0.
The p-value and significance level are provided from the questions .
General rule of P value : When a P value is less than or equal to the significance level, you reject the null hypothesis.
