Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 76.6 Mbps. The complete list of 50 data speeds has a mean of x overbarequals18.39 Mbps and a standard deviation of sequals31.86 Mbps. a. What is the difference between carrier's highest data speed and the mean of all 50 data speeds?
b. How many standard deviations is that [the difference found in part (a)]?
c. Convert the carrier's highest data speed to a z score.
d. If we consider data speeds that convert to z scores between minus2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant?

Step-by-step explanation:

Hello!

The variable of interest is

X: data speed of a particular smartphone carrier in one airport.

n= 50 airports

X[bar]= 18.39 Mbps

S= 31.86 MBps

The highest speed recorded X= 76.6 Mbps

a.

The difference between the highest speed and the mean is:

X-X[bar]= 76.6-18.39= 58.21 Mbps

b.

To calculate how many standard deviations away from the mean the highest speed is, you have to divide it by the standard deviation:

c.

The standard normal distribution is tabulated. Any value of any random variable X with normal distribution can be "converted" by subtracting the variable from its mean and dividing it by its standard deviation. This is exactly what you did in items a. and b., so the Z score is

Z= 1.82

d.

-2 ≤ Z ≤ 2⇒ the calculated Z value is within both limits, so the carrier's highest data speed is not significant.

I hope this helps!

a

step-by-step explanation:

a

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.