A random sample of size n = 82 is taken from a population with mean = 15.6 and s
ID: 3383014 • Letter: A
Question
A random sample of size n = 82 is taken from a population with mean = 15.6 and standard deviation = 5 use Z chart
Calculate the expected value and the standard error for the sampling distribution of the sample mean. (Negative values should be indicated by a minus sign. Round "expected value" to 1 decimal place and "standard deviation" to 4 decimal places.)
Expected value :
Standard error :
What is the probability that the sample mean is less than 16?
What is the probability that the sample mean falls between 16 and 15?
Explanation / Answer
The expected value is the same as the population mean,
Expected value = -15.6 [ANSWER]
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Meanwhile,
Standard error = sigma/sqrt(n) = 5/sqrt(82) = 0.55215763 [ANSWER]
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LESS THAN -16:
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = -16
u = mean = -15.6
n = sample size = 82
s = standard deviation = 5
Thus,
z = (x - u) * sqrt(n) / s = -0.72
Thus, using a table/technology, the left tailed area of this is
P(z < -0.72 ) = 0.2358 [ANSWER]
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BETWEEN -16 AND -15:
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = -16
x2 = upper bound = -15
u = mean = -15.6
n = sample size = 82
s = standard deviation = 5
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -0.72
z2 = upper z score = (x2 - u) * sqrt(n) / s = 1.09
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.2358
P(z < z2) = 0.8621
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.6263 [ANSWER]
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