In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed with a mean of 54 inches, and standard deviation of 8 inches. What is the probability that the height of a randomly chosen child is between 38.9 and 61 inches

77.98% probability that the height of a randomly chosen child is between 38.9 and 61 inches

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:

What is the probability that the height of a randomly chosen child is between 38.9 and 61 inches

This is the pvalue of Z when X = 61 subtracted by the pvalue of Z when X = 38.9. So

X = 61

has a pvalue of 0.8092

X = 38.9

has a pvalue of 0.0294

0.8092 - 0.0294 = 0.7798

77.98% probability that the height of a randomly chosen child is between 38.9 and 61 inches

135 is your answer

step-by-step explanation:

try 180 times three then divide that number by 4 (sides) to get your answer. im glad to : )

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: