A class has average height of 68 inches and a standard division of 2 inches what height do you need to be taller than 90% of the class please round your answer to the nearest thousandth.

65.426

Step-by-step explanation:

Given that:

Mean height (m) = 68 inches

Standard deviation (s) = 2 inches

Height required to be taller than 90% of the class :

P(Z > x) = 0.9

The corresponding Zscore is - 1.282 (Z probability calculator)

Using the Zscore relation :

Zscore = (x - m) / s

-1.282 = (x - 68) / 2

-1.282 * 2 = x - 68

-2.564 = x - 68

-2.564 + 68 = x

x = 65.436

i'm pretty sure it's 477,585,296, but perhaps my math is incorrect.

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.