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?

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