Question

Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 57 hours and a standard deviation of 3.5 hours. With this information, answer the following questions. (a) What proportion of light bulbs will last more than 61 hours? (b) What proportion of light bulbs will last 52 hours or less? (c) What proportion of light bulbs will last between 57 and 61 hours? (d) What is the probability that a randomly selected light bulb lasts less than 46 hours? (a) The proportion of light bulbs that last more than 61 hours is (Round to four decimal places as needed.)

Answer 1

Solution:

We are given

µ = 57

σ = 3.5

Part a

We have to find P(X>61) = 1 - P(X<61)

Z = (X - µ)/σ

Z = (61 - 57)/3.5

Z = 1.142857

P(Z<1.142857) = P(X<61) = 0.873451

(You can find this probability by using z-table/excel/Ti-84/83 calculator/software/etc.)

P(X>61) = 1 - P(X<61)

P(X>61) = 1 - 0.873451

P(X>61) = 0.126549

Required probability = 0.1265

Part b

We have to find P(X≤52)

Z = (X - µ)/σ

Z = (52 - 57)/3.5

Z = -1.42857

P(Z<-1.42857) = P(X≤52) = 0.076564

(You can find this probability by using z-table/excel/Ti-84/83 calculator/software/etc.)

P(X≤52) = 0.076564

Required probability = P(X≤52) = 0.0766

Part c

We have to find P(57<X<61)

P(57<X<61) = P(X<61) - P(X<57)

P(X<61) = 0.873451

(From part a)

Z = (X - µ)/σ

Z = (57 - 57)/3.5

Z = 0

P(Z<0) = P(X<57) = 0.50000

(You can find this probability by using z-table/excel/Ti-84/83 calculator/software/etc.)

P(57<X<61) = P(X<61) - P(X<57)

P(57<X<61) = 0.873451 - 0.50000

P(57<X<61) = 0.373451

Required probability = 0.3735

Part d

We have to find P(X<46)

Z = (X - µ)/σ

Z = (46 - 57)/3.5

Z = -3.14286

P(Z<-3.14286) = P(X<46) = 0.000837

(You can find this probability by using z-table/excel/Ti-84/83 calculator/software/etc.)

Required probability = 0.0008