Let X be the mean of a random sample of size n = 48 from the uniform distributio
ID: 3386762 • Letter: L
Question
Let X be the mean of a random sample of size n = 48 from the uniform distribution in the interval (0, 2). Approximate the probability P (0.9 < X < 1.1) using the Central Limit Theorem. Let X be the mean of a random sample of size n = 48 from the uniform distribution in the interval (0, 2). Approximate the probability P (0.9 < X < 1.1) using the Central Limit Theorem. Let X be the mean of a random sample of size n = 48 from the uniform distribution in the interval (0, 2). Approximate the probability P (0.9 < X < 1.1) using the Central Limit Theorem.Explanation / Answer
Note that here,
a = lower fence of the distribution = 0
b = upper fence of the distribution = 2
Thus, the mean, variance, and standard deviations are
u = mean = (b + a)/2 = 1
s^2 = variance = (b -a)^2 / 12 = 0.333333333
s = standard deviation = sqrt(s^2) = 0.577350269
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 0.9
x2 = upper bound = 1.1
u = mean = 1
n = sample size = 48
s = standard deviation = 0.577350269
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -1.2
z2 = upper z score = (x2 - u) * sqrt(n) / s = 1.2
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.11506967
P(z < z2) = 0.88493033
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.76986066 [ANSWER]
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