A factory produces ball bearings whose diameters are meant to have a mean of 3 mm. Suppose that the actual sizes are normally distributed with a mean of 3 mm and a standard deviation of 0.2 mm. The overseers took a
random sample of n = 100 ball bearings to see if their mean diameter was significantly different than the
target. The mean diameter of the ball bearings in the sample was c = 2.96 mm.
To see how likely a sample like theirs was to occur by random chance alone, the factory overseers performed a
simulation. They simulated 75 samples of n = 100 diameters from a normal population with a mean of 3 mm
and standard deviation of 0.2 mm. They recorded the mean of the diameters in each sample. Here are the
samnle means from their 75 samples:

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.  

probably the last option

step-by-step explanation: