X~N(60,14). Suppose that you form random samples of 25 from this distribution. L
ID: 3149877 • Letter: X
Question
X~N(60,14). Suppose that you form random samples of 25 from this distribution. Let X be the random variable of averages Let (EX) or (sum)X be the random variable of sums.
A.) Find the 20th percentile. (round 4 decimal places)
B.) Give the distribution of SumX. Fill in blank N(1500,______)
C.) Find the minimum Value for the upper quartile for sum X. Round your answer to two decimal places.
D.) Find the probability of P(1400S SumX 1550)= _____________.
Please show all work and if there is a Z score could you explain how to put it into a TI83 calculator? I am struggling here!
Thank you in advance!
Explanation / Answer
a)
First, we get the z score from the given left tailed area. As
Left tailed area = 0.2
Then, using table or technology,
z = -0.841621234
As x = u + z * s / sqrt(n)
where
u = mean = 60
z = the critical z score = -0.841621234
s = standard deviation = 14
n = sample size = 25
Then
x = critical value = 57.64346055 [ANSWER]
***************************
b)
Note that for samples of size n,
SumX~(n*u, sigma*sqrt(n))
Hence,
SumX~(25*60, 14*sqrt(25))
Thus,
SumX~(1500, 70) [ANSWER]
***************************
c)
The upper quartile is the 75th percentile.
First, we get the z score from the given left tailed area. As
Left tailed area = 0.75
Then, using table or technology,
z = 0.67448975
As x = u + z * s / sqrt(n)
where
u = mean = 60
z = the critical z score = 0.67448975
s = standard deviation = 14
n = sample size = 25
Then
x = critical value = 61.8885713 [ANSWER]
****************************
d)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 1400
x2 = upper bound = 1550
u = mean = 1500
s = standard deviation = 70
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -1.428571429
z2 = upper z score = (x2 - u) / s = 0.714285714
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.076563726
P(z < z2) = 0.762474738
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.685911012 [ANSWER]
***************************
You can get the left tailed area of a z score using normcdf(-999999, z, 0, 1).
You can get the z score from a left tailed area using InvNorm(left tailed area, 0, 1).
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.