Twenty-five percent of all automobiles undergoing an emissions inspection at a c
ID: 3262693 • Letter: T
Question
Twenty-five percent of all automobiles undergoing an emissions inspection at a certain inspection station fail the inspection. (Round all answers to three decimal places.)
(a) Among 18 randomly selected cars, what is the probability that at most 6 fail the inspection?
(b) Among 18 randomly selected cars, what is the probability that between 6 and 12 (inclusive) fail to pass inspection?
(c) Among 25 randomly selected cars, what is the mean value of the number that pass inspection, and what is the standard deviation of the number that pass inspection?
Mean =
Standard deviation =
(d) What is the probability that among 25 randomly selected cars, the number that pass is within 1 standard deviation of the mean value?
Explanation / Answer
Question a)
This is Binomial Distribution:
n = 18
p = 0.25
p (x) = C (n,x) *p^x * (1-p)^(n-x)
p ( x<= 6 ) = P ( x = 0 ) + P ( x = 1 ) + P ( x = 2 ) + P ( x = 3 ) + P ( x = 4 ) + P ( x = 5 ) + P ( x = 5 )
x
Formula
P (x)
0
C(18,0) * 0.25^0 * 0.75^18
0.0056
1
C(18,1) * 0.25^1 * 0.75^17
0.0338
2
C(18,2) * 0.25^2 * 0.75^16
0.0958
3
C(18,3) * 0.25^3 * 0.75^15
0.1704
4
C(18,4) * 0.25^4 * 0.75^14
0.2130
5
C(18,5) * 0.25^5 * 0.75^13
0.1988
6
C(18,6) * 0.25^6 * 0.75^12
0.1436
P (x<= 6 ) = 0.0056 + 0.0338 + 0.0958 + 0.1704 + 0.2130 + 0.1988 + 0.1436
= 0.8610
Answer: 0.8610
Question b)
P ( 6 <= x <= 12 )
= P ( x = 6 ) + P ( x = 7 ) + P ( x = 8 ) + P ( x = 9 ) + P ( x = 10 ) + P ( x = 11 ) + P (x = 12 )
6
C(18,6) * 0.25^6 * 0.75^12
0.1436
7
C(18,7) * 0.25^7 * 0.75^11
0.0820
8
C(18,8) * 0.25^8 * 0.75^10
0.0376
9
C(18,9) * 0.25^9 * 0.75^9
0.0139
10
C(18,10) * 0.25^10 * 0.75^8
0.0042
11
C(18,11) * 0.25^11 * 0.75^7
0.0010
12
C(18,12) * 0.25^12 * 0.75^6
0.0002
= 0.1436 + 0.0820 + 0.0376 + 0.0139 + 0.0042 + 0.0010 + 0.0002
= 0.2825
Answer: 0.2825
Question c)
n = 25
p = 0.75
Mean = np = 25*0.75 = 18.75
Standard deviation = sqrt (npq) = sqrt (25*0.75*(1-0.75)) = 2.1651
Mean = 18.75
Standard deviation = 2.1651
Question d)
Within 1 standard deviation
18.75 – 2.16 = 17
18.75 + 2.16 = 21
P ( 17 < x < 21 ) = P ( x = 18 ) + P ( x = 19 ) + P ( x = 20 )
x
Formula
P (x)
18
C(25,18) * 0.75^18 * 0.25^7
0.1654
19
C(25,19) * 0.75^19 * 0.25^6
0.1828
20
C(25,20) * 0.75^20 * 0.25^5
0.1645
P ( 17 < x < 21 ) = P ( x = 18 ) + P ( x = 19 ) + P ( x = 20 )
= 0.1654 + 0.1828 + 0.1654
= 0.5127
Answer: 0.5127
x
Formula
P (x)
0
C(18,0) * 0.25^0 * 0.75^18
0.0056
1
C(18,1) * 0.25^1 * 0.75^17
0.0338
2
C(18,2) * 0.25^2 * 0.75^16
0.0958
3
C(18,3) * 0.25^3 * 0.75^15
0.1704
4
C(18,4) * 0.25^4 * 0.75^14
0.2130
5
C(18,5) * 0.25^5 * 0.75^13
0.1988
6
C(18,6) * 0.25^6 * 0.75^12
0.1436
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