Question

The average monthly rental for a 2-bedroom apartment in Austin is $950. Assume that the population mean is $950 and the population standard deviation is $200.

1. What is the probability that a simple random sample of 50 2-bedroom apartments will provide a sample mean monthly rental within +/- $100 of the population mean??

2. What is the probability that a simple random sample of 50 2-bedroom apartments will provide a sample mean monthly rental within +/- $25 of the population mean?

Answer #1

1)

Here, μ = 950, σ = 28.2843, x1 = 850 and x2 = 1050. We need to compute P(850<= X <= 1050). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ

z1 = (850 - 950)/28.2843 = -3.54

z2 = (1050 - 950)/28.2843 = 3.54

Therefore, we get

P(850 <= X <= 1050) = P((1050 - 950)/28.2843) <= z <=
(1050 - 950)/28.2843)

= P(-3.54 <= z <= 3.54) = P(z <= 3.54) - P(z <=
-3.54)

= 0.9998 - 0.0002

= 0.9996

2)

Here, μ = 950, σ = 28.2843, x1 = 925 and x2 = 975. We need to compute P(925<= X <= 975). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ

z1 = (925 - 950)/28.2843 = -0.88

z2 = (975 - 950)/28.2843 = 0.88

Therefore, we get

P(925 <= X <= 975) = P((975 - 950)/28.2843) <= z <=
(975 - 950)/28.2843)

= P(-0.88 <= z <= 0.88) = P(z <= 0.88) - P(z <=
-0.88)

= 0.8106 - 0.1894

= 0.6212

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