Question

A random sample of size n = 152 is taken from a population of size N = 3,300 with mean μ = −71 and variance σ² = 112. [You may find it useful to reference the z table.]

a-1. Is it necessary to apply the finite population correction factor?

• Yes

• No

a-2. Calculate the expected value and the standard error of the sample mean. (Negative values should be indicated by a minus sign. Round "standard error" to 2 decimal places.)

b. What is the probability that the sample mean is between −73 and −69? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

c. What is the probability that the sample mean is greater than −70? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

Answer 1

a-1)

No,

It is not necessary to apply the finite population correction factor.

a-2)

Expected value = = -71

Standard error = / sqrt(n)

= sqrt(112) / sqrt(152)

= 0.86

b)

Using central limit theorem,

P( < x = P( Z < x - / ( / sqrt(n) ) )

Therefore,

P(-73 < < -69) = P( < -69) - P( < -73)

= P( Z < -69 + 71 / 0.86) - P( Z < -73 + 71 / 0.86)

= P( Z < 2.33) - P( Z < -2.33)

= P( Z < 2.33) - P( Z < -2.33)

= 0.9901 - 0.0099

= 0.9802

c)

P( > -70) = P( Z > -70 + 71 / 0.86)

= P( Z > 1.16)

= 1 - P( Z < 1.16)

= 1 - 0.877

= 0.123

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