Question

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?

Answer 1

Answer:

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:

$\frac{X-X[bar]}{S} = \frac{58.21}{31.86}= 1.82$

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!