Question

Name: 1. The time required for a citizen to complete the 2000 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. What proportion of the citizens will require less than one hour to complete the census? 2. The time required for a citizen to complete the 2000 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. The slowest 10% of the citizens would need how many minutes to complete the form? 3. Sampling distributions describe the distribution of a) parameters. b) statistics. c) both parameters and statistics. d) neither parameters nor statistics 4. The standard error of the mean a. is never larger than the standard deviation of the population. b. decreases as the sample size increases. c. measures the variability of the mean from sample to sample. d. All of the above. 5. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. What is the standard error for the sample mean? Show your work a. 0.029 b. 0.050 c. 0.091 d. 0.120 6. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. What is the probability that the sample mean will be between 0.99 and 1.01 centimeters?

Answer 1

1.

Filling time, X ~ Normal(μ, σ)

μ = 40
σ = 10

P(X < 60) = P{(X - μ)/σ < (60 - 40)/10}

= P(Z < 2)

= 0.9773 [use table]

So, 0.97.73% of the citizens will require less than 60 minutes.

2.

The slowest 10% is at the right-most side of the distribution.

So, F(Z) = 1 - 0.1 = 0.90

or, Z = 1.28 [use table]

So, X = μ + Z.σ = 40 + 1.28*10 = 52.8

So, the slowest 10% will need 52.8 minutes or more for filling the form.

3.

(b) Statistics

4.

(d) All of the above

5.

σ = 0.10 cm
n = 12

So, Std. Error (σ_x̄) = σ / √n = 0.10/√12 = 0.029

6.

μ_x̄ = 1.0
σ_x̄ = 0.029

P(0.99 < x̄ < 1.01)

= P(x̄ < 1.01) - P(x̄ < 0.99)

= P(z < (1.01 - 1)/0.029) - P(z < (0.99 - 1)/0.029)

= P(z < 0.345) - P(z < -0.345)

= 0.635 - 0.365 [use table]

= 0.270