Given a normal distribution with mean 100 and standard deviation 10, find the area under the normal curve from the mean to 118.8.

The area under the normal curve from the mean to 118.8. is 0.47

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean and standard deviation , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

Find the area under the normal curve from the mean to 118.8.

This is the pvalue of Z when X = 118.8 subtracted by the pvalue of Z when X = 100.

X = 118.8

has a pvalue of 0.97

X = 100

has a pvalue of 0.5

0.97 - 0.5 = 0.47

The area under the normal curve from the mean to 118.8. is 0.47

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