Suppose X is the breaking strength (newtons) of a material, and X is normally distributed withmean 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 theprobability 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 morethan 820?

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 newtonsStandard Deviation, σ = 10 newtonsWe are given that the distribution of  breaking strength is a bell shaped distribution that is a normal distribution. Formula: [tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex] a) P( breaking strength of at least 772 newtons) [tex]P(x \geq 772)[/tex] [tex]P( x \geq 772) = P( z \geq \displaystyle\frac{772 - 800}{10}) = P(z \geq -2.8)[/tex] [tex]= 1 - P(z <-2.81)[/tex] Calculation the value from standard normal z table, we have,  [tex]P(x \geq 772) = 1 - 0.003 = 0.997 = 99.7\%[/tex]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) [tex]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\%[/tex] [tex]P(772 \leq x \leq 820) = 97.4\\%[/tex]0.974  is the probability that this material has a breaking strength of at least 772 but not more  than 820.