Question

Suppose that a brand of lightbulb lasts on average 1730 hours with a standard deviation of 257 hours. Assume the life of the lightbulb is normally distributed. Calculate the probability that a particular bulb will last from 1689 to 2267 hours?

Answer 1

P [ 1689 ≤ X ≤ 2267 ] = 54,88 %

Explanation:

Normal Distribution

Mean μ₀ = 1730

Standard Deviation σ = 257

We need to calculate z scores for the values 1689 and 2267

We apply formula for z scores

z = ( X - μ₀ ) /σ

X = 1689 then

z = (1689 - 1730)/ 257 ⇒ z = - 41 / 257

z = - 0.1595

And from z table we get for z = - 0,1595

We have to interpolate

- 0,15 0,4364

- 0,16 0,4325

Δ = 0.01 0.0039

0,1595 - 0,15 = 0.0095

By rule of three

0,01 0,0039

0,0095 x ?? x = 0.0037

And 0,4364 - 0.0037 = 0,4327

Then P [ X ≤ 1689 ] = 0.4327 or P [ X ≤ 1689 ] = 43,27 %

And for the upper limit 2267 z score will be

z = ( X - 1730 ) / 257 ⇒ z = 537 / 257

z = 2.0894

Now from z table we find for score 2.0894

We interpolate and assume 0.9815

P [ X ≤ 2267 ] = 0,9815

Ths vale already contains th value of P [ X ≤ 1689 ] = 0.4327

Then we subtract to get 0,9815 - 0,4327 = 0,5488

Finally

P [ 1689 ≤ X ≤ 2267 ] = 0,5488 or P [ 1689 ≤ X ≤ 2267 ] = 54,88 %