if u can help part a,b and c. plz 1a. A light fixture hold two light bulbs. Bulb

Question

if u can help part a,b and c. plz

1a. A light fixture hold two light bulbs. Bulb 1 has a lifetime that is normally distributed with a mean of 1000 hours and a standard deviation of 50 hours. Buib 2 has higher mean life (1200 hours) but Higher variability ( 150 hours). Assume independent failure modes for the bulbs, estimate the probability that bulb 1 lasts longer than bulb 2. A random variable X follows Uniform distribution with parameters 4 and 8. Estimate the probability of the event: Pl X5). Does it satisfy lack of memory property? 1b. 1c. Specification for a bolt important for safe operation of a machine requires that the tensile strength be at least 25 kN (kilo Newton). The manufacturing process output follows a Normal distribution. It is known that 5% of the process output falls below 25.5 kN and 5% output record strength over 27 kN. Estimate the percentage of process output that meets the specification, 1d. A random variable X follows Exponential distribution with mean -10. Find the Inter Quartile Range (IQR) of the distribution.

Explanation / Answer

Question 1(a) Two light bulbs. Let say the life of first bulb is X and life of the second bulb is Y.

Here X ~ N(1000, 50) and Y ~ N(1200, 150)

So, we have to find the probability of event when life of bulb 1 (which is X ) is greater than life of bulb 2 (which is Y).

that , written mathematically,

Pr( X -Y > 0) = ?

Here E(X-Y) = 1000 - 1200 = -200 hours

Variance of (X-Y) = Var(X - Y) = Var(X) + Var(Y) = 50 2 + 1502 = 25000

Stadad deviaiton of (X-Y) = Sqrt(25000) = 158.114 hours

so Pr( X -Y > 0) = NORMAL (X -Y > 0 ; -200; 158.114) = 1- NORM (X -Y < 0 ; -200; 158.114)

Z = (0 + 200)/ 158.114 = 1.265

so Pr( X -Y > 0) = 1- Pr (Z < 1.265) = 1 - 0.8971 = 0.1029

(b) Here the distribution is unifrom distribution.

so let say X is the random variable having this distribution.

so,

f(x) = 1/(8-4) = 1/4

CDF ...F(x) = (x-4)/4 ; 4 < x < 8

so Pr( X < 6.5 l x > 5) = Pr( 5 < x < 6.5)/ Pr(X ? 5) = F(6.5) - F(5) = 1/4 [ (6.5-4) - (5 -4)] / [1 - (5-4)/4) ] = 1/4 [2.5 - 1] = 0.375 / 0.75 = 0.5

Yes, it satisfy lack of memoryless property.

(c) Minimum tensile strength required = 25 kN

Here let say X is the tensile strength of a random manufacturing output. Let say process mean is and standard deviation is .

Pr(X < 25.5 kN ;   ; ) = 0.05

Pr(X > 27 kN ;   ; ) = 0.05

so mean process strength = (25.5 + 27)/2 = 26.25 kN

Here Z value for p - vlaue = 0.05 is Z = 1.645

1.645 = (27 - 26.25)/

= 0.75/ 1.645 = 0.456 kN

so now we have to find the percentage of process output that meets the specification.

Pr(X > 25 kN ) = NORM (X > 25 kN ; 26.25 ; 0.456) = 1 - NORM( X < 25 kN ; 26.25 ; 0.456)

Z = (25 - 26.25)/ 0.456 = -2.7412

so Pr(X > 25 kN ) = 1 - Pr(Z < -2.7412) = 1 - 0.003 = 0.997

so 99.70 % process output is out of specification limit.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.