Suppose that birth weights are normally distributed with a mean of 3466 grams and a standard deviation of 546 grams. Babies above the 95th percentile have a condition called macrosomia (also known as "big baby syndrome"). What birth weight marks the 95th percentile?

A birth weight of 4364.17 grams marks the 95th percentile.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

What birth weight marks the 95th percentile?

This is X when Z has a pvalue of 0.95. So it is X when Z = 1.645.

A birth weight of 4364.17 grams marks the 95th percentile.

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x