The fracture strength of a certain type of manufactured glass is normally distributed with a mean of 509 MPa with a standard deviation of 17 MPa. (a) What is the probability that a randomly chosen sample of glass will break at less than 509 MPa

0.5 = 50% probability that a randomly chosen sample of glass will break at less than 509 MPa

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 509 MPa with a standard deviation of 17 MPa.

This means that

What is the probability that a randomly chosen sample of glass will break at less than 509 MPa?

This is the p-value of Z when X = 509. So

has a p-value of 0.5

0.5 = 50% probability that a randomly chosen sample of glass will break at less than 509 MPa

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b