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!

75 tickets have been sold so far.

step-by-step explanation:

in this question it is clearly mentioned that 45 children's tickets have been sold and this number is 60% of the total tickets sold.

now let's assume the total number of tickets sold be x.

then 60% of x = 45

⇒

⇒ x(.6) = 45

by dividing with .6 on both the sides of the equation

⇒ x = 75

therefore 75 tickets have been sold so far.

you can't solve x because it is 2 linear equation and you only have one equation