Answer:
Shannon run 1.5 km more on Friday than on Saturday.
Step-by-step explanation:
From the given table
- Distance run on Friday = 4/2 = 2 km
- Distance run on Saturday = 1/2 = 0.5 km
- In order to run how many more kilometers Shannon run on Friday than on Saturday, we need to subtract the distance run on Saturday from the distance run on Friday.
i.e.
Friday run - Saturday run = 2 - 0.5
= 1.5 km
Thus, Shannon run 1.5 km more on Friday than on Saturday.
You multiply .65 times 100 and you get 65 which is 65% of 100
Answer:(9,2) I assume I don’t really understand the question
Step-by-step explanation:
Answer:
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the grade points avergae of a population, and for this case we know the following properties
Where and
The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).
So we can find the z score for the value of X=3.44 in order to see how many deviations above or belowe we are from the mean like this:
So the value of 3.44 is 2 deviations above from the mean, so then we know that the percentage between two deviations from the mean is 95% and on each tail we need to have (100-95)/2 = 2.5% , because the distribution is symmetrical, so based on this we can conclude that: