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2 years ago

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Construct a 95% confidence interval for the population mean, µ. Assume the population has a normal distribution. A sample of 25

randomly selected students has a mean test score of 81.5 with a standard deviation of 10.2. a. (66.35, 69.89) b. (56.12, 78.34) c. (77.29, 85.71) d. (87.12, 98.32)

1 answer:

bixtya [17]2 years ago

6 0

Answer:C. (77.29, 85.71)

Step-by-step explanation:

We want to determine a 95% confidence interval for the mean test score of randomly selected students.

Number of sample, n = 25

Mean, u = 81.5

Standard deviation, s = 10.2

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean +/- z ×standard deviation/√n

It becomes

81.5 +/- 1.96 × 10.2/√25

= 81.5 +/- 1.96 × 2.04

= 81.5 +/- 3.9984

The lower end of the confidence interval is 81.5 - 3.9984 =77.5016

The upper end of the confidence interval is 81.5 + 3.9984 =85.4984

Therefore, the correct option is

C. (77.29, 85.71)

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To see the steps to the diagonal form see the step-by-step explanation. The solution to the system is $x = -\frac{1}{9}$ , $y= -\frac{1}{9}$ , $z= \frac{4}{9}$ and $w = \frac{7}{9}$

Step-by-step explanation:

Gauss elimination method consists in reducing the matrix to a upper triangular one by using three different types of row operations (this is why the method is also called row reduction method). The three elementary row operations are:

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To solve the system using the Gauss elimination method we need to write the augmented matrix of the system. For the given system, this matrix is:

$\left[\begin{array}{cccc|c}1 & 1 & 1 & 1 & 1 \\1 & 1 & 0 & -1 & -1 \\-1 & 1 & 1 & 2 & 2 \\1 & 2 & -1 & 1 & 0\end{array}\right]$

For this matrix we need to perform the following row operations:

$R_2 - 1 R_1 \rightarrow R_2$ (multiply 1 row by 1 and subtract it from 2 row)
$R_3 + 1 R_1 \rightarrow R_3$ (multiply 1 row by 1 and add it to 3 row)
$R_4 - 1 R_1 \rightarrow R_4$ (multiply 1 row by 1 and subtract it from 4 row)

$R_2 \leftrightarrow R_3$ (interchange the 2 and 3 rows)

$R_2 / 2 \rightarrow R_2$ (divide the 2 row by 2)
$R_1 - 1 R_2 \rightarrow R_1$ (multiply 2 row by 1 and subtract it from 1 row)
$R_4 - 1 R_2 \rightarrow R_4$ (multiply 2 row by 1 and subtract it from 4 row)
$R_3 \cdot ( -1) \rightarrow R_3$ (multiply the 3 row by -1)
$R_2 - 1 R_3 \rightarrow R_2$ (multiply 3 row by 1 and subtract it from 2 row)
$R_4 + 3 R_3 \rightarrow R_4$ (multiply 3 row by 3 and add it to 4 row)
$R_4 / 4.5 \rightarrow R_4$ (divide the 4 row by 4.5)

After this step, the system has an upper triangular form

The triangular matrix looks like:

$\left[\begin{array}{cccc|c}1 & 0 & 0 & -0.5 & -0.5 \\0 & 1 & 0 & -0.5 & -0.5\\0 & 0 & 1 & 2 & 2 \\0 & 0 & 0 & 1 & \frac{7}{9}\end{array}\right]$

If you later perform the following operations you can find the solution to the system.

$R_1 + 0.5 R_4 \rightarrow R_1$ (multiply 4 row by 0.5 and add it to 1 row)
$R_2 + 0.5 R_4 \rightarrow R_2$ (multiply 4 row by 0.5 and add it to 2 row)
$R_3 - 2 R_4 \rightarrow R_3$ (multiply 4 row by 2 and subtract it from 3 row)

After this operations, the matrix should look like:

$\left[\begin{array}{cccc|c}1 & 0 & 0 & 0 & -\frac{1}{9} \\0 & 1 & 0 & 0 & -\frac{1}{9}\\0 & 0 & 1 & 0 & \frac{4}{9} \\0 & 0 & 0 & 1 & \frac{7}{9}\end{array}\right]$

Thus, the solution is:

$x = -\frac{1}{9}$ , $y= -\frac{1}{9}$ , $z= \frac{4}{9}$ and $w = \frac{7}{9}$

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