A manufacturer claims that a certain brand of VCR player has an average life expectancy of 5 years and 6 months with a standard deviation of 1 year and 6 months. Assume that the life expectancy is normally distributed.14. Based on the information above, when selecting one VCR player at random what is the probability of its life expectancy being greater than seven years

15.87% probability of its life expectancy being greater than seven years

Step-by-step explanation:

Problems of normally distributed samples are solved using 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 problem, we have that:

Life expectancy of 5 years and 6 months

6 months is half an year, so

Standard deviation of 1 year and 6 months.

So

Probability of its life expectancy being greater than seven years

This is 1 subtracted by the pvalue of Z when X = 7.

has a pvalue of 0.8413

1 - 0.8413 = 0.1587

15.87% probability of its life expectancy being greater than seven years

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.

answer:

it is general formula that is used to write value to its excet position

a11 a12 a13

a21 a22 a23

it is the example of order 2 by 3 matrix in this

a11 means value occur in 1 st row 1st column

a12 means value occurs in 1st row 2nd column

a13 means value occur in 1st row and 3rd column

a21 means value occurs in 2nd row and 1st column

a22 means value occurs in 2nd row and in 2nd column

a23 means value occurs in 2nd row and in 3rd column