Answer:
Step-by-step explanation:
Researchers measured the data speeds for a particular smartphone carrier at 50 airports.
The highest speed measured was 76.6 Mbps.
n= 50
X[bar]= 17.95
S= 23.39
a. What is the difference between the carrier's highest data speed and the mean of all 50 data speeds?
If the highest speed is 76.6 and the sample mean is 17.95, the difference is 76.6-17.95= 58.65 Mbps
b. How many standard deviations is that [the difference found in part (a)]?
To know how many standard deviations is the max value apart from the sample mean, you have to divide the difference between those two values by the standard deviation
Dif/S= 58.65/23.39= 2.507 ≅ 2.51 Standard deviations
c. Convert the carrier's highest data speed to a z score.
The value is X= 76.6
Using the formula Z= (X - μ)/ δ= (76.6 - 17.95)/ 23.39= 2.51
d. If we consider data speeds that convert to z scores between minus−2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant?
The Z value corresponding to the highest data speed is 2.51, considerin that is greater than 2 you can assume that it is significant.
I hope it helps!
<u></u><u>The correct answer is 47.5%, or 0.475.</u>
Explanation:
The empirical rule states that in any normal distribution:
68% of data will fall within 1 standard deviation of the mean;
95% of data will fall within 2 standard deviations of the mean; and
99.7% of data will fall within 3 standard deviations of the mean.
The mean is 500 and the standard deviation is 100. This means that 700 is 2 standard deviations away from the mean:
(700-500)/100=200/100=2.
We know that 95% of data will fall within 2 standard deviations from the mean. However, included in the 95% is data less than the mean and greater than the mean. Since we are only concerned with the scores from 500 to 700, we only want the half that is greater than the mean:
95/2 = 47.5%, or 0.475.
Answer:
B
Step-by-step explanation:
3(2+x)
distribute
=6+3x
74.4 is the answer.
hope this helps.
