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.

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option