Question 1200150: Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 15 days and a standard deviation of 6 days. Let X be the number of days for a randomly selected trial. Round all answers to 4 decimal places where possible.
a. What is the distribution of X? X ~ N(
(15,6)
Correct
b. If one of the trials is randomly chosen, find the probability that it lasted at least 17 days.
Correct: 0.3694
c. If one of the trials is randomly chosen, find the probability that it lasted between 14 and 18 days.
0.2576
Correct
d. 80% of all of these types of trials are completed within how many days? (Please enter a whole number)
* I need help with letter B on finding at least in normal distribution and letter D on a TI-84 Calculator
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (b)
On your TI84 calculator, press the button labeled "2nd" in the upper left corner.
Then press the VARS key to bring up the stats distribution menu.
The function we're after is normalcdf which finds the area under the normal distribution curve.
The template is normalcdf(L, U, mu, sigma)
L = lower boundary
U = upper boundary
mu = mean
sigma = standard deviation
In the case of part (b), we have
L = 17
U = 9999, or any other very large value to represent positive infinity
mu = 15
sigma = 6
This means you would type in normalcdf(17,9999,15,6)
Your calculator should produce the approximate result of 0.3694414037 which rounds to 0.3694 when rounding to four decimal places.
Answer to part (b): 0.3694
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Part (d)
We're looking for a value of k such that
P(X < k) = 0.80
we want to know the cut-off point between the lower 80% and the upper 20%
The function we want is called invNorm
This can be found in the same menu as normalcdf. It's in the 3rd slot.
The template for this function is
invNorm(p, mu, sigma)
where,
p = area under the normal curve between 0 and 1
mu = mean
sigma = standard deviation
In this case,
p = 0.80
mu = 15
sigma = 6
So you should type in invNorm(0.80, 15, 6)
The calculator will produce a result of roughly 20.0497274
That rounds to 20 when rounding to the nearest whole number.
So we can say that
P(X < 20) = 0.80
approximately.
About 80% of the trials finish within 20 days (i.e. 20 days or fewer).
Answer to part (d) is 20
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