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

Answer:

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

Step-by-step 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 %