The tensile strength of paper is modeled by a normal distribution with a mean of 35 pounds per square inch and a standard deviation of 2 pounds per square inch. Required: a. What is the probability that the strength of a sample is less than 38 lb/in²? b. If the specifications require the tensile strength to exceed 30 lb/in², what proportion of the samples is scrapped?

Answer:a) the probability that the strength of a sample is less than 38 lb/in² is approximately 0.9332b) approximately 0.62% of the sample are scrappedStep-by-step explanation:a) What is the probability that the strength of a sample is less than 38 lb/in²?lets X rep the paper tensile strength in pounds per square inch.the probability that the strength of the sample is less than 38 lb/in²i.e P( X < 38 )so assuming that X is normally distributed with u = 35 lb/in² and α = 2 lb/in²,so the probability expression P( X < 38 ) can be standardized as P( X < 38 ) = P( (X-35)/2  < (38-35)/2)  = P( X < 1.5 )   from the cumulative standard normal distribution table, we find P( X < 1.5 )  =  0.9332so P( X < 38 ) = P( X < 1.5 )  = 0.9332therefore the probability that the strength of a sample is less than 38 lb/in² is approximately 0.9332b)If the specifications require the tensile strength to exceed 30 lb/in², what proportion of the samples is scrapped?this can be represented as P( X < 30 )By standardization the probability expression  P( X < 30 ) can be transformed as; P( X < 30 ) = P( (X-35)/2 < (30-35)/2)  =  P( z < -2.5 ) from the cumulative standard normal distribution table, we find P( X < -2.5 )  =  0.0062 so P( X < 30 ) = P( X < -2.5 )  = 0.0062therefore, approximately 0.62% of the sample are scrapped