Question

The length of life of oil-drilling bits depends upon the types of rock and soil that the drill encounters, but it is estimated that the mean length of life is 55 hours. An oil exploration company purchases drill bits that have a length of life that is approximately normally distributed with mean equal to 55 hours and standard deviation equal to 9.1 hours.
What proportion of the company’s drill bits will have to be replaced after more than 90 hours of use?

Answer 1

Answer:

"0.9999" is the correct solution.

Step-by-step explanation:

The given values in the question are:

Mean,

= 55

Standard deviation,

= 9.1

Now,

⇒ $z=\frac{X-mean}{sd}$

By putting the values, we get

⇒    $=\frac{90-55}{9.1}$

⇒    $=\frac{35}{9.1}$

⇒    $=3.8461$

hence,

$P(X$

= $P(Z$

= $NORMSDIST(3.8461)$

then the value will be:

= $0.9999$