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Answer :
Certainly! Let's break down each part of the question and provide detailed, step-by-step solutions.
### Question 1
Problem Statement:
X and Y Companies each claim their mean annual repair bill is Birr 290 with a standard deviation of Birr 50. We want to find the probability that the difference between the sample means [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex] (where [tex]\( x_1 \)[/tex] is the mean repair bill of a sample of 100 cars from X Company and [tex]\( x_2 \)[/tex] is the corresponding mean from Y Company) exceeds Birr 5.
Solution:
1. Calculate the standard error (SE) of the difference between two means:
[tex]\[
SE = \sqrt{\frac{\sigma_X^2}{n_X} + \frac{\sigma_Y^2}{n_Y}}
\][/tex]
Here, [tex]\(\sigma_X = \sigma_Y = 50\)[/tex], [tex]\(n_X = n_Y = 100\)[/tex].
[tex]\[
SE = \sqrt{\frac{50^2}{100} + \frac{50^2}{100}} = \sqrt{\frac{2500}{100} + \frac{2500}{100}} = \sqrt{25 + 25} = \sqrt{50} \approx 7.07
\][/tex]
2. Calculate the z-score corresponding to a difference of 5 Birr:
[tex]\[
z = \frac{5}{SE} = \frac{5}{7.07} \approx 0.707
\][/tex]
3. Use the standard normal distribution to find the corresponding probability:
The probability that [tex]\( x_1 - x_2 \)[/tex] exceeds 5 is:
[tex]\[
P(Z > 0.707) \approx 0.2398
\][/tex]
So, the probability that the difference in sample means exceeds Birr 5 is approximately 0.2398.
### Question 2
Problem Statement:
Two populations have the following characteristics:
- Population 1: Mean = 57, Standard Deviation = 12
- Population 2: Mean = 25, Standard Deviation = 6
Independent samples of size 36 are taken from each population.
Solution:
i) Expected value of the difference in sample means ([tex]\( x_1 - x_2 \)[/tex]):
[tex]\[
E(x_1 - x_2) = \mu_1 - \mu_2 = 57 - 25 = 32
\][/tex]
ii) Standard deviation of the distribution of [tex]\( x_1 - x_2 \)[/tex]:
[tex]\[
\sigma_{x_1 - x_2} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} = \sqrt{\frac{12^2}{36} + \frac{6^2}{36}} = \sqrt{\frac{144}{36} + \frac{36}{36}} = \sqrt{4 + 1} = \sqrt{5} \approx 2.236
\][/tex]
So, the expected value of the difference in sample means is 32, and the standard deviation of the distribution is approximately 2.236.
### Question 3
Problem Statement:
The heights of a certain group of boys (given in meters) are: 1.50, 1.60, 1.60, 1.70, 1.80, and 1.80. We are to consider all possible samples of size 3 drawn with replacement and calculate certain metrics.
Solution:
a) Mean height of the population:
To find the population mean, sum all heights and divide by the number of data points.
[tex]\[
\mu = \frac{1.50 + 1.60 + 1.60 + 1.70 + 1.80 + 1.80}{6} \approx 1.6667
\][/tex]
b) Standard deviation of the population:
The standard deviation is given by:
[tex]\[
\sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{N}}
\][/tex]
Using the provided results:
[tex]\[
\sigma \approx 0.1106
\][/tex]
c) Construct the sampling distribution of means:
We consider all possible samples of size 3:
[tex]\[
\text{Samples: } [1.50, 1.60, 1.60, 1.70, 1.80, 1.80] \text{ (repeated for combinations)}
\][/tex]
Consider the means of all these combinations. This produces a series of means as shown in the data provided.
d) Mean of the sampling distribution:
As given, the mean of the sampling distribution of means is the same as the population mean due to the central limit theorem.
[tex]\[
\mu_{\bar{x}} = \mu = 1.6667
\][/tex]
So, the processed calculations show:
- The mean height of the population is approximately 1.6667 meters.
- The standard deviation of the population is approximately 0.1106 meters.
- The sampling distribution consists of a series of means calculated from all possible samples of size 3.
- The mean of this sampling distribution is the same as the population mean, approximately 1.6667 meters.
### Question 1
Problem Statement:
X and Y Companies each claim their mean annual repair bill is Birr 290 with a standard deviation of Birr 50. We want to find the probability that the difference between the sample means [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex] (where [tex]\( x_1 \)[/tex] is the mean repair bill of a sample of 100 cars from X Company and [tex]\( x_2 \)[/tex] is the corresponding mean from Y Company) exceeds Birr 5.
Solution:
1. Calculate the standard error (SE) of the difference between two means:
[tex]\[
SE = \sqrt{\frac{\sigma_X^2}{n_X} + \frac{\sigma_Y^2}{n_Y}}
\][/tex]
Here, [tex]\(\sigma_X = \sigma_Y = 50\)[/tex], [tex]\(n_X = n_Y = 100\)[/tex].
[tex]\[
SE = \sqrt{\frac{50^2}{100} + \frac{50^2}{100}} = \sqrt{\frac{2500}{100} + \frac{2500}{100}} = \sqrt{25 + 25} = \sqrt{50} \approx 7.07
\][/tex]
2. Calculate the z-score corresponding to a difference of 5 Birr:
[tex]\[
z = \frac{5}{SE} = \frac{5}{7.07} \approx 0.707
\][/tex]
3. Use the standard normal distribution to find the corresponding probability:
The probability that [tex]\( x_1 - x_2 \)[/tex] exceeds 5 is:
[tex]\[
P(Z > 0.707) \approx 0.2398
\][/tex]
So, the probability that the difference in sample means exceeds Birr 5 is approximately 0.2398.
### Question 2
Problem Statement:
Two populations have the following characteristics:
- Population 1: Mean = 57, Standard Deviation = 12
- Population 2: Mean = 25, Standard Deviation = 6
Independent samples of size 36 are taken from each population.
Solution:
i) Expected value of the difference in sample means ([tex]\( x_1 - x_2 \)[/tex]):
[tex]\[
E(x_1 - x_2) = \mu_1 - \mu_2 = 57 - 25 = 32
\][/tex]
ii) Standard deviation of the distribution of [tex]\( x_1 - x_2 \)[/tex]:
[tex]\[
\sigma_{x_1 - x_2} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} = \sqrt{\frac{12^2}{36} + \frac{6^2}{36}} = \sqrt{\frac{144}{36} + \frac{36}{36}} = \sqrt{4 + 1} = \sqrt{5} \approx 2.236
\][/tex]
So, the expected value of the difference in sample means is 32, and the standard deviation of the distribution is approximately 2.236.
### Question 3
Problem Statement:
The heights of a certain group of boys (given in meters) are: 1.50, 1.60, 1.60, 1.70, 1.80, and 1.80. We are to consider all possible samples of size 3 drawn with replacement and calculate certain metrics.
Solution:
a) Mean height of the population:
To find the population mean, sum all heights and divide by the number of data points.
[tex]\[
\mu = \frac{1.50 + 1.60 + 1.60 + 1.70 + 1.80 + 1.80}{6} \approx 1.6667
\][/tex]
b) Standard deviation of the population:
The standard deviation is given by:
[tex]\[
\sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{N}}
\][/tex]
Using the provided results:
[tex]\[
\sigma \approx 0.1106
\][/tex]
c) Construct the sampling distribution of means:
We consider all possible samples of size 3:
[tex]\[
\text{Samples: } [1.50, 1.60, 1.60, 1.70, 1.80, 1.80] \text{ (repeated for combinations)}
\][/tex]
Consider the means of all these combinations. This produces a series of means as shown in the data provided.
d) Mean of the sampling distribution:
As given, the mean of the sampling distribution of means is the same as the population mean due to the central limit theorem.
[tex]\[
\mu_{\bar{x}} = \mu = 1.6667
\][/tex]
So, the processed calculations show:
- The mean height of the population is approximately 1.6667 meters.
- The standard deviation of the population is approximately 0.1106 meters.
- The sampling distribution consists of a series of means calculated from all possible samples of size 3.
- The mean of this sampling distribution is the same as the population mean, approximately 1.6667 meters.
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