Question

An average light bulb manufactured at The Lightbulb Company lasts and average of 300 days, with a standard deviation of 50 days. Suppose the lifespan of a light bulb from this company is normally distributed. (a) What is the probability that a light bulb from this company lasts less than 210 days? More than 330 days?(b) What is the probability that a light bulb from this company lasts between 280 and 380 days?(c) How would you characterize the lifespan of the light bulbs whose lifespans are among the shortest 2% of all bulbs made by this company?(d) If a pack of 6 light bulbs from this company are purchased, what is the probability that exactly 4 of them last more than 330 days?

Answer 1

Answer:

a). 0.03593, 0.27425

(bl 0.0703

(c) the lifespan of such bulb is less than 210 days

(d) 0.0298.

Step-by-step explanation:

Given that U = 300, sd= 50

Z = X-U/sd

(a) We find the probability

Pr(X<210) = Pr(Z<(210-300)/50)

= Pr(Z<-1.8)

= 0.03593

Pr(X >330) = Pr(Z >(330-300)/50))

= Pr(Z> 0.6)

= 1- PR(Z<0.6) = 1 - 0.72575

Pr(X>330)= 0.27425

(b) Pr(280< X< 330) = Pr( 280 < Z< 380)

= Pr([280-300/50] < Z < [380-300/50])

= Pr(0.4 < Z < 1.6)

= Pr(Z< 1.6) - Pr(Z < 0.4)

=0.72575 - 0.65542

= 0.07033

= 0.0703

(c) the lifespan of such bulbs is less than 210 days

(d) given n = 6, x = 4 ,

Pr(X>330) = 0.27425 = p

q = 1-p = 0.72575

Pr(X=4) = 6C4 × (0.27425)⁴ ×(0.72575)²

Pr(X=4) = 10 × 0.005657 × 0.52671

Pr(X= 4) = 0.029795

Pr(X= 4) = 0.0298