One solution to the problem below is 4. What is the other solution? y²

A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal

GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean income = $\frac{\sum X}{n}$ = $4.70

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = $1.83

n = sample of parking meters = 10

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.262 < $t_9$ < 2.262) = 0.95 {As the critical value of t at 9 degrees of

freedom are -2.262 & 2.262 with P = 2.5%}

P(-2.262 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.262) = 0.95

P( $-2.262 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu$ < $2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $4.70-2.262 \times {\frac{1.83}{\sqrt{10} } }$ , $4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } }$ ]

= [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error = $2.262 \times {\frac{s}{\sqrt{n} } }$

= $2.262 \times {\frac{1.83}{\sqrt{10} } }$ = 1.31

4 0

3 years ago

Plz help me with this question

Alex17521 [72]

Your answer for this question will be the second one on the list

5 0

3 years ago

Read 2 more answers

Based on the graph, which inequality is correct for a number that is to the right of -3?

Firdavs [7]

Answer:

4 > −3

Step-by-step explanation:

To find a number to the right of -3, it must be bigger than -3

The open part of the inequality faces the bigger number.

4 is bigger than -3

4>-3

3 0

3 years ago

You paint 1/2 wall in 1/4hour At that rate how long will it take you to paint one wall

3241004551 [841]

Answer:

half hour

Step-by-step explanation:

Do 1 fourth times two

8 0

3 years ago

A vector with magnitude 9 points in a direction 190 degrees counterclockwise from the positive x axis. Write in component form

Over [174]

Answer:

$\vec{v}= \text{ or } \approx$

Step-by-step explanation:

Component form of a vector is given by $\vec{v}=$ , where $i$ represents change in x-value and $j$ represents change in y-value. The magnitude of a vector is correlated the Pythagorean Theorem. For vector $\vec{v}=$ , the magnitude is $||v||=\sqrt{i^2+j^2$ .

190 degrees counterclockwise from the positive x-axis is 10 degrees below the negative x-axis. We can then draw a right triangle 10 degrees below the horizontal with one leg being $i$ , one leg being $j$ , and the hypotenuse of the triangle being the magnitude of the vector, which is given as 9.

In any right triangle, the sine/sin of an angle is equal to its opposite side divided by the hypotenuse, or longest side, of the triangle.

Therefore, we have:

$\sin 10^{\circ}=\frac{j}{9},\\j=9\sin 10^{\circ}$

To find the other leg, $i$ , we can also use basic trigonometry for a right triangle. In right triangles only, the cosine/cos of an angle is equal to its adjacent side divided by the hypotenuse of the triangle. We get:

$\cos 10^{\circ}=\frac{i}{9},\\i=9\cos 10^{\circ}$

Verify that $(9\sin 10^{\circ})^2+(9\cos 10^{\circ})^2=9^2\:\checkmark$

Therefore, the component form of this vector is $\vec{v}=\boxed{}\approx \boxed{}$

6 0

3 years ago

One solution to the problem below is 4. What is the other solution? y² - 16 = 0