The distribution lies within one of the standard of deviation of the mean so <span>68% </span>
The distribution lies within two of the standard of deviations of the mean so 95%
The distribution lies within three of the standard of deviations of the mean so 99.7%
Hi!
To compare this two sets of data, you need to use a t-student test:
You have the following data:
-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph
-Wednesday n2=20; </span>x̄2=56,3 mph; s2=4,4 mph
You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.
To calculate the degrees of freedom you need to use the following equation:
≈34
The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10
So, as the calculated value is higher than the critical tabulated one,
we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.
Answer: The answer is 100,
Step-by-step explanation:
You would get 10$ for the first week, 20 for the second week, 30 for the third week, and 40 for the fourth week. If you add that up all together you get 100$
You can draw a picture and you can find your answer but the answer is 1 1/3 by the way