Great Free Practice Problems
2 answers:
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Answer:
a) 59.98
b) 2.99
c) 2.99
d) Significantly High
Step-by-step explanation:
Part a)
Highest speed measured = x = 75.6 Mbps
Average/Mean speed = = 15.62 Mbps
Standard Deviation = s = 20.03 Mbps
We need to find the difference between carrier's highest data speed and the mean of all 50 data speeds i.e. x -
x - = 75.6 - 15.62 = 59.98 Mbps
Thus, the difference between carrier's highest data speed and the mean of all 50 data speeds is 59.98 Mbps
Part b)
In order to find how many standard deviations away is the difference found in previous part, we divide the difference by the value of standard deviation i.e.
This means, the difference is 2.99 standard deviations or in other words we can say, the Carrier's highest data speed is 2.99 standard deviations above the mean data speed.
Part c)
A z score tells us that how many standard deviations away is a value from the mean. We calculated the same in the previous part. Performing the same calculation in one step:
The formula for the z score is:
Using the given values, we get:
Thus, the Carriers highest data is equivalent to a z score of 2.99
Part d)
The range of z scores which are neither significantly low nor significantly high is -2 to + 2. The z scores outside this range will be significant.
Since, the z score for carrier's highest data speed is 2.99 which is well outside the given range, i.e. greater than 2, we can conclude that the carrier's highest data speed is significantly higher.