The following is the probability mass function for the number of times a certain computer program will malfunction: 0 1 2 3 4 5 p(x) | 0.05 0.28 0.42 0.15 0.08 0.02 (a) what is the probability that the computer program will malfunction more than 3 times? (b) compute e(x), e(x2), and v(x). e(x) = e(x2) = v(x) = (c) suppose y, the time in minutes to fix malfunctions, relates to the number of times the program malfunctions, by the function: y = 9*x . what is the expected time in minutes needed to fix malfunctions in the program? d) what is the variance of the time in minutes to fix the malfunctions?

a)P(X>3)=0.1

b)

E(X)=1.99

E(X²)=5.09

V(X)=1.13

c) E(Y)=17.91

d) V(Y)=81*V(X)=91.5

Step-by-step explanation:

a)

P(X>3)=P(X=4)+P(X=5)=0.08+0.02=0.1

b)

E(X)=∑x*p(x)=0*0.05+1*0.28+2*0.42+3*0.15+4*0.08+5*0.02=1.99

E(X²)=∑x²*p(x)=0*0.05+1*0.28+4*0.42+9*0.15+16*0.08+25*0.02=5.09

V(X)=E(X²)-(E(X))²=5.09-3.96=1.13

c)

E(Y)=9*E(X)=9*1.99=17.91

d)

V(Y)=9²*V(X)=81*V(X)=81*1.13=91.5

hey! so 1 (25) out of 4 (100) is 25 so your answer is 25%. it's like quarters!

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;