The average daily rainfall for the past week in the town of Hope Cove is normally distributed, with a mean rainfall of 2.1 inches and a standard deviation of 0.2 inches. If the distribution is normal, what percent of data lies between 1.9 inches and 2.3 inches of rainfall? a) 95% b) 99.7% c) 34% d) 68%

D

Step-by-step explanation:

We calculate the z-score for each

Mathematically;

z-score = (x-mean)/SD

z1 = (1.9-2.1)/0.2 = -1

z2 = (2.3-2.1)/0.2 = 1

So the proportion we want to calculate is;

P(-1<x<1)

We use the standard score table for this ;

P(-1<x<1) = P(x<1) -P(x<-1) = 0.68269 which is approximately 68%

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.