Question

Suppose X is the breaking strength (newtons) of a material, and X is normally distributed with

mean 800 newtons and standard deviation of 10 newtons.
a) Suppose we need this material to have breaking strength of at least 772 newtons. What is the
probability that the breaking strength is at least 772 newtons.
b) What is the probability that this material has a breaking strength of at least 772 but not more
than 820?

Answer 1

a) 0.997 is the probability that the breaking strength is at least 772 newtons.

b) 0.974 is the probability that this material has a breaking strength of at least 772 but not more than 820

Explanation:

We are given the following information in the question:

Mean, μ = 800 newtons

Standard Deviation, σ = 10 newtons

We are given that the distribution of breaking strength is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

a) P( breaking strength of at least 772 newtons)

`P(x \geq 772)`

`P( x \geq 772) = P( z \geq \displaystyle(772 - 800)/(10)) = P(z \geq -2.8)`

`= 1 - P(z <-2.81)`

Calculation the value from standard normal z table, we have,

`P(x \geq 772) = 1 - 0.003 = 0.997 = 99.7\%`

0.997 is the probability that the breaking strength is at least 772 newtons.

b) P( breaking strength of at least 772 but not more than 820)

`P(772 \leq x \leq 820) = P(\displaystyle(772 - 800)/(10) \leq z \leq \displaystyle(820-800)/(10)) = P(-2.8 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2.8)\\= 0.977 - 0.003 = 0.974 = 97.4\%`

`P(772 \leq x \leq 820) = 97.4\\%`

0.974 is the probability that this material has a breaking strength of at least 772 but not more than 820.