A population has a mean of 300 and a standard deviation of 60. Suppose a sample of size 100 is selected and is used to estimate . Use z-table.
What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?
What is the probability that the sample mean will be within +/- 12 of the population mean (to 4 decimals)?
Mean, = 300
Standard deviation, = 60
Sample size, n = 100
Sample mean, = 300
Standard error, =
=
= 6
P(sample mean will be within +/- 5 of the population mean) = P(-5/ < Z < 5/)
= P(-5/6 < Z < 5/6)
= P(-0.83 < Z < 0.83)
= P(Z < 0.83) - P(Z < -0.83)
= 0.7969 - 0.2033
= 0.5936
P(sample mean will be within +/- 12 of the population mean) = P(-12/ < Z < 12/)
= P(-12/6 < Z < 12/6)
= P(-2 < Z < 2)
= P(Z < 2) - P(Z < -2)
= 0.9772 - 0.0228
= 0.9544
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