Question

QUESTION 19 A manufacturer of bike racks for cars claims that the assembly time x for a particular model is normally distributed with a mean of 1 hour and a standard deviation of 0.10 hours. Find the probability that it takes at most 1.1 hours to assemble a bike rack of this model. QUESTION 20 The standard error of a statistic used as an estimator of a population parameter is: A. The same value as the population standard deviation. B. The same value of the population variance C. The standard deviation of the sampling distribution of the statistic. D. The variance of the sampling distribution of the statistic. QUESTION 21 Suppose an actual census showed that 20% of the households in Michigan have incomes in excess of $60,000. Assume that a random sample of 500 households in Michigan is taken. Then, the standard error of the sampling distribution of sample proportion of households who have incomes in excess of $60,000 will be: A. 0.0179 B. 0.1600 C. 0.0003 D. 0.0256

Answer 1

Solution:

Question 19)

Given, X follows Normal distribution with,

  $\mu$ = 1

$\sigma$ = 0.10

Now ,

P[at most 1.1]

= P[X $\leq$ 1.1]

= P[(X - $url=https://latex.codecogs.com/gif.latex?%5Cmu&t=1596163916&ymreqid=f4764063-0d40-e2b0-1cde-f00053012f00&sig=PrkMvSl1tOa6cBMo7G19mg--~D$ )/$url=https://latex.codecogs.com/gif.latex?%5Csigma&t=1596163916&ymreqid=f4764063-0d40-e2b0-1cde-f00053012f00&sig=UUwRWH102YhAW8Ogu4PJHg--~D$$\leq$  (1.1 - $url=https://latex.codecogs.com/gif.latex?%5Cmu&t=1596163916&ymreqid=f4764063-0d40-e2b0-1cde-f00053012f00&sig=PrkMvSl1tOa6cBMo7G19mg--~D$ )/$url=https://latex.codecogs.com/gif.latex?%5Csigma&t=1596163916&ymreqid=f4764063-0d40-e2b0-1cde-f00053012f00&sig=UUwRWH102YhAW8Ogu4PJHg--~D$]

= P[Z $\leq$  (1.1 - 1)/0.10]

= P[Z $\leq$ 1.00]

= 0.8413 ... ( use z table)

Answer is

0.8413

Question 20.

Correct option is

C. The standard deviation of the sampling distribution of the statistic.

Question 21.

Given ,

p = 20% = 0.20(population proportion)

n = 500 (sample size)

Let $url=https://latex.codecogs.com/gif.latex?%5Chat%20p&t=1596163916&ymreqid=f4764063-0d40-e2b0-1cde-f00053012f00&sig=pT4wg6C5NHe1iKvYzbn_bg--~D$ be the sample proportion.

$url=https://latex.codecogs.com/gif.latex?%5Ctherefore&t=1596163916&ymreqid=f4764063-0d40-e2b0-1cde-f00053012f00&sig=skvzcEQbJodJ8MOwbwWjEg--~D$ The sampling distribution of $url=https://latex.codecogs.com/gif.latex?%5Chat%20p&t=1596163916&ymreqid=f4764063-0d40-e2b0-1cde-f00053012f00&sig=pT4wg6C5NHe1iKvYzbn_bg--~D$ is approximately normal with

mean = $\mu_{\hat p}$ =  p = 0.20

SD = $\sigma_{\hat p}$   =   $url=https://latex.codecogs.com/gif.latex?%5Csqrt%7Bp%281-p%29/n%7D&t=1596163916&ymreqid=f4764063-0d40-e2b0-1cde-f00053012f00&sig=3dAFoyLfyTKzYc5GHFH_HA--~D$ =   = 0.0179

(0.2011-0.200/500

Answer is

A. 0.0179