Question

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

Answer 1

Answer:

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    

Step-by-step 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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

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

$P(x \geq 772)$

$P( x \geq 772) = P( z \geq \displaystyle\frac{772 - 800}{10}) = P(z \geq -2.8)$

$= 1 - P(z$

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)

$P(772 \leq x \leq 820) = P(\displaystyle\frac{772 - 800}{10} \leq z \leq \displaystyle\frac{820-800}{10}) = P(-2.8 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2.8)\\= 0.977 - 0.003 = 0.974 = 97.4\%$

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

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