theyycraveenaee
17.03.2020 •
Mathematics
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?
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Ответ:
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
Ответ:
9/2
Step-by-step explanation:
4.5 = 4 1/2 = 9/2