The sample std. dev. will be (14 inches) / sqrt(49), or (14 inches) / 7, or 2 inches.
Find the z score for 93.8 inches:
93.8 inches - 91.0 inches 2.8 inches
z = ------------------------------------- = ----------------- = 1.4
2 inches 2 inches
Now find the area under the standard normal curve to the left of z = +1.4.
My calculator returns the following:
normalcdf(-100,1.4) = 0.919. This is the probability that the mean annual precipitation during those 49 years will be less than 93.8 inches.
Answer:
Z(2w-2)=24
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The data for resort A shows more consistency because a larger interquartile range such as the one for resort B, shows more variation. This means that the snowfall for resort A is more likely to be close to the median.
Just did this on edg. :)
Answer:
Step-by-step explanation:
<u>We can observe that:</u>
- The difference of the second and third numbers is equal to 17
- The difference of the first and fourth numbers is equal to 13
<u>Therefore the missing pair will be:</u>
- 20+17 = 37 and 28 + 13 = 41
Correct choice is A
None of the offered choices is correct.
The correct equation is
.. a = (2s -2ut)/t^2 . . . . . . . . parentheses are required
