Question

The average lifetime of a light bulb is 3000 hours with a standard deviation of 696 hours. A simple random sample of 36 bulbs is taken.

What is the probability that the average life in the sample will be greater than 3219.24?

Note: respond in decimal form, for example, if your solution is 87.1234%, you should enter in 0.871234.

Answer 1

Solution :

Given, X follows Normal distribution with,

   = 3000

   = 696

A sample of size n = 36 is taken from this population.

Let be the mean of sample.

The sampling distribution of the is approximately normal with

Mean = = 3000

SD =    = 696/​36 = 116

Find P( > 3219.24)

= P[( - )/ >  (3219.24 - )/]

= P[ ( - )/ > (3219.24 - 3000)/116 ]

= P[Z > 1.89 ]

= 1 - P[Z < 1.89]

= 1 - 0.9706 ..........( use z table)

= 0.0294

P( > 3219.24) = 0.0294