A train leaves at 08.43 and arrives at its destination at 09.23. If the train travelled 73 km, what was it's average speed in km
1 answer:
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If you are looking to write an inequality equation that models this situation, then the answer would be: . The "at most" signals to use the less than or equal to sign.
Answer:
a) P(-z_0.025 < Z < z_0.025)
For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:
"=NORM.INV(0.025,0,1)"
"=NORM.INV(0.025,0,1)"
And for this case the two values are :
b) P(-z_{\alpha/2} < Z < z_{\alpha/2})
For this case we want a quantile that accumulates of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:
"=NORM.INV(alpha/2,0,1)"
"=NORM.INV(alpha/2,0,1)"
c) For this case we want to find a value of z that satisfy:
P(Z > z_alpha) = 0.05.
And we can use the following excel code:
"=NORM.INV(0.95,0,1)"
And we got
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
Part a
P(-z_0.025 < Z < z_0.025)
For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:
"=NORM.INV(0.025,0,1)"
"=NORM.INV(0.025,0,1)"
And for this case the two values are :
Part b
P(-z_{\alpha/2} < Z < z_{\alpha/2})
For this case we want a quantile that accumulates of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:
"=NORM.INV(alpha/2,0,1)"
"=NORM.INV(alpha/2,0,1)"
Part c
For this case we want to find a value of z that satisfy:
P(Z > z_alpha) = 0.05.
And we can use the following excel code:
"=NORM.INV(0.95,0,1)"
And we got