Respuesta :
Answer:
a. The probability of the random variable is as follows
P(1) = 0.475
P(2) = 0.396
P(3) = 0.110
P(4) = 0.015
P(5) = 0.004 -
b. The probability that a student takes the SAT more than one time is calculated as 0.525
c. The probability that a student takes the SAT three or more times is calculated as 0.129
d. The Expected value E(x) is calculated as 1.7
This means that the number of students that took the SAT test are approximately 1.7 times
e.. The variance = 0.579
Step-by-step explanation:
a.
Calculating total number of students.
Total = 721,769 + 601,325 + 166,736 + 22,299 + 6,730
Total = 1,518,859
The probability of the random variable is calculated as
Number of observations/Number of total outcomes
So, the probability of the random variable is as follows
P(1) = 721,769/1,518,859
P(1) = 0.47520474250736901845398
P(1) = 0.475 --- Approximated
P(2) = 601,325/1,518,859
P(2) = 0.39590574240268517354145
P(2) = 0.396
P(3) = 166,736/1,518,859
P(3) = 0.10977714192034942018976
P(3) = 0.110 --- Approximated
P(4) = 22,299/1,518,859
P(4) = 0.01468141545726100974481
P(4) = 0.015 ---- Approximated
P(5) = 6,730/1,518,859
P(5) = 0.00443095771233537806998
P(5) = 0.004 --- Approximated
b. The probability that a student takes the SAT more than one time is calculated as
P(2) +.....+P(5)
= 0.396 + 0.110 + 0.015 + 0.004
= 0.525
Alternatively, it can be calculated as 1 - P(1)
= 1 - 0.475
= 0.525
c. The probability that a student takes the SAT three or more times is calculated as
P(3) + P(4) + P(5)
= 0.110 + 0.015 + 0.004
= 0.129
d. What is the expected value of the number of times the SAT is taken?
What is your interpretation of the expected value?
The Expected value E(x) is calculated as
∑xf(x)
X --- f(x) -------- xf(x) ------ x - u -------- (x-u)² ---- (x-u)²f(x)
P(1) = 0.475 ---- 0.475 --- -0.677 ---- 0.457 ----- 0.218
P(2) = 0.396 --- 0.792 ---- 0.323 ---- 0.104 ----- 0.412
P(3) = 0.110 ---- 0.330 --- 1.323 ------ 1.750 ----- 0.192
P(4) = 0.015 ---- 0.06 ---- 2.323 ---- 5.400 ----- 0.792
P(5) = 0.004 --- 0.02 ---- 3.323 ----- 11.041 ----- 0.049
∑xf(x) = 1.677 = 1.7 ---- Approximated
This means that the number of students that took the SAT test are approximately 1.7 times
E. What is the variance
Var (x) = ∑(x-u)²f(x)
Var(x) = 0.579
Answer / Explanation
(a) Num of time(x) Num of students Probability
1 721769 721769 / 1518859 = 0.4752
2 601325 601325 / 1518859 = 0.3959
3 166736 166736 / 1518859 = 0.1098
4 22299 22299 / 1518859 = 0.0147
5 6730 6730 / 1518859 = 0.0044
(B) probability that a student took the SAT exam more than one time:
f ( x > 1 ) = 1 - f (x ≤ 1)
= 1 - f ( x = 1)
= 1 - 0.4752
= 0.5248
(C) Probability that a student took the SAT three or more time:
= f ( x ≥ 3 ) = f ( x = 3 ) + f ( x = 4 ) + f ( x = 5 )
= 0.1098 + 0.0147 + 0.0044
= 0.1289
(D) Expected value of x is E (x) = μ
E (x) = ∑ x f (x)
= 1 x 0.4752 + 2 x 0.3959 + 3 x 0.1098 + 4 x 0.0147 + 5 x 0.0044
0.4752 + 0.7918 + 0.3294 + 0.0588 + 0.022
= 1.6772
Therefore, it can be concluded that the expected number of time the student took SAT exam is approximately 2 time.
(E) Variance of x = б²
б² = ∑ ( x - μ )² f (x)
= ( 1 - 1.6772)² x 0.4752 +2 ( 2 - 1.6772 ) ² x 0.3959 + ( 3 - 1.6772)² x 0.1098 + ( 4 - 1.6772)² x 0.0147 + ( 5 - 1.6772)² x 0.0044
= ( - 0.6672)² x 0.4752 + ( 0.3228)² x 0.3959 + ( 1.3228)² x 0.1098 + (2.3228)² x 0.0044
= 0.4452 x 0.4752 + 0.1042 x 0. 3935 + 1.7498 x 0.1098 + 5.3954 x 0.0147 + 11.0410 + 0.0044
= 0.2163 + 0.0412 + 0.1921 + 0.0793 + 0.0486
= 0.5775
Recalling that standard deviation of x = б
Therefore,
б = √б²
= √0.5775
= 0.76
