Please help me I am on a test
2 answers:
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Answer:
So on this case the 98% confidence interval would be given by (3.767;4.233)
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
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".
represent the sample mean
population mean (variable of interest)
represent the population standard deviation
n=100 represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
(1)
Since the Confidence is 0.98 or 98%, the value of and
, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.01,0,1)".And we see that
Now we have everything in order to replace into formula (1):
So on this case the 98% confidence interval would be given by (3.767;4.233)
V1 - velocity first train
v2 - velocity second train
v2 > v1
v2 - v1 = 17 mph
We know, that:
![s_1+s_2=210 \ [miles] \\ \\ t_1=t_2=2h](https://tex.z-dn.net/?f=s_1%2Bs_2%3D210%20%5C%20%5Bmiles%5D%20%5C%5C%20%5C%5C%20t_1%3Dt_2%3D2h)
So:
NOw we've got simple system of equations:
![+\begin{cases} v_2-v_1=17 \\ v_2+v_1=105\end{cases} \\ \\ 2v_2=122 \qquad /:2 \\ \\ v_2=61 \qquad [mph] \\ \\ v_2-v_1=17 \\ \\ 61-v_1=17 \\ \\ v_1=44](https://tex.z-dn.net/?f=%2B%5Cbegin%7Bcases%7D%20v_2-v_1%3D17%20%5C%5C%20v_2%2Bv_1%3D105%5Cend%7Bcases%7D%20%5C%5C%20%5C%5C%202v_2%3D122%20%5Cqquad%20%2F%3A2%20%5C%5C%20%5C%5C%20v_2%3D61%20%5Cqquad%20%5Bmph%5D%20%5C%5C%20%5C%5C%20v_2-v_1%3D17%20%5C%5C%20%5C%5C%2061-v_1%3D17%20%5C%5C%20%5C%5C%20v_1%3D44)
Velocities of these trains are 61mph and 44mph
v2 - velocity second train
v2 > v1
v2 - v1 = 17 mph
We know, that:
So:
NOw we've got simple system of equations:
Velocities of these trains are 61mph and 44mph