The answer is 3.6. Hope this helps!
Answer:
Mean = 1.42
Variance = 0.58
Step-by-step explanation:
Given: X denote the number of luxury cars sold in a given day, and Y denote the number of extended warranties sold.
Also, joint probability function of X and Y are given.
To find:
mean and variance of X
Solution:
From the given joint probability function of X and Y,
Mean of X:
Variance of X:
![var(X)=E\left [ X^2 \right ]-\left ( E\left [ X \right ] \right )^2\\=\frac{31}{12}-\left ( \frac{17}{12} \right )^2\\=\frac{31}{12}-\frac{289}{144}\\=\frac{372-289}{144}\\=\frac{83}{144}\\=0.58](https://tex.z-dn.net/?f=var%28X%29%3DE%5Cleft%20%5B%20X%5E2%20%5Cright%20%5D-%5Cleft%20%28%20E%5Cleft%20%5B%20X%20%5Cright%20%5D%20%5Cright%20%29%5E2%5C%5C%3D%5Cfrac%7B31%7D%7B12%7D-%5Cleft%20%28%20%5Cfrac%7B17%7D%7B12%7D%20%5Cright%20%29%5E2%5C%5C%3D%5Cfrac%7B31%7D%7B12%7D-%5Cfrac%7B289%7D%7B144%7D%5C%5C%3D%5Cfrac%7B372-289%7D%7B144%7D%5C%5C%3D%5Cfrac%7B83%7D%7B144%7D%5C%5C%3D0.58)
Answer:
27
Step-by-step explanation:
<u><em>First, you would do 7+2 to get 9 and then multiply it by itself to get 81. Next, you divide 81 by 3 to get 27.</em></u>
Answer:0.29
Step-by-step explanation:
An average of six cell phone thefts is reported in San Francisco per day. This means our mean value, u = 6
For poisson distribution,
P(x=r) = (e^-u×u^r)/r!
probability that four cell phones will be reported stolen tomorrow=
P(x=4)= (e^-6×6^4)/4!
= (0.00248×1296)/4×3×2×1
= 3.21408/24=
0.13392
P(x=5)= (e^-6×6^5)/5!
= (0.00248×7776)/5×4×3×2×1
= 19.28448/120
= 0.1607
probability that four or five cell phones will be reported stolen tomorrow
= P(x=4) + P(x=5)
= 0.13392 + 0.1607
= 0.294624
Approximately 0.29
Answer:
P(4≤x≤7) = 2/3
Step-by-step explanation:
We'll begin by obtaining the sample space (S) i.e possible outcome of rolling both dice at the same time. This is illustrated below:
1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6
3,1 3,2 3,3 3,4 3,5 3,6
4,1 4,2 4,3 4,4 4,5 4,6
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6
Adding the outcome together, the sample space (S) becomes:
2 3 4 5 6 7
3 4 5 6 7 8
4 5 6 7 8 9
5 6 7 8 9 10
6 7 8 9 10 11
7 8 9 10 11 12
Next, we shall obtain the event of 4≤x≤7. This is illustrated below:
4 5 6 7
4 5 6 7
4 5 6 7
4 5 6 7
4 5 6 7
4 5 6 7
Finally, we shall determine P(4≤x≤7). This can be obtained as follow:
Element in the sample space, n(S) = 36
Element in 4≤x≤7, n(4≤x≤7) = 24
Probability of 4≤x≤7, P(4≤x≤7) = ?
P(4≤x≤7) = n(4≤x≤7) / nS
P(4≤x≤7) = 24/36
P(4≤x≤7) = 2/3
