Suppose that the lifetimes of light bulbs are approximately normally distributed

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 thes intormaton answer the tolowing auesions (a) What proportion of light bultbs will last more than 60 hours? (b) What proportion of light bulbs will last 51 hours or less? (c) What proportion of light bulbs will last between 57 and 62 hours? (d) What is the probability that a randomly selected light bulb lasts less than 45 hours? (a) The proportion of light bulbs that last more than 60 hours is Round to four decimal places as needed) Enter your answer in the answer box and then click Check Ansvr amaining Type here to search

Explanation / Answer

We have the relation X = + Z

= 57 = 3.5

(a) X = 60

=> 60 = 57 + 3.5Z

=> 3.5Z = 3

=> Z = 3/3.5

=> Z = 0.8571

Using the value in the Z-table, we have 0.8023 or 80%.

Therefore, approximately 20% bulbs will last more than 60 hours.

(b) X = 51

=> 51 = 57 + 3.5Z

=> 3.5Z = -6

=> Z = -6/3.5

=> Z = -1.714

From the table, the value is 0.436 or 44%.

Thus approximately 44% bulbs will last 51 hours or less.

(c) X = 57

=> 57 = 57 + 3.5Z

=> Z = 0

The value is 0.5 or 50%.

X = 62

62 = 57 + 3.5Z

=> Z = 5/3.5 = 1.483

From the table, the value is 0.93 or 93%

Number of bulbs lasting between 57 and 62 hours = 93 - 50 = 43%.

(d) X = 45

=> 45 = 57 + 3.5Z

=> Z = -12/3.5

=> Z = -3.429

From the table, the value is 0.0003 .

The probability of a light bulb lasting less than 45 hours = 0.0003

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