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1 year ago

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Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United Sta

tes is $165,000. This distribution follows the normal distribution with a standard deviation of $40,000. If we select a random sample of 50 households, what is the standard error of the mean

1 answer:

Molodets [167]1 year ago

4 0

Answer: $5656.85

Step-by-step explanation:

The standard error of the mean measures the dispersion of sample means about the population mean .

It is also known as the standard deviation of its sampling distribution.

Formula : $SE=\dfrac{\sigma}{\sqrt{n}}$ , where $\sigma$ = population standard deviation.

n= sample size.

As per given , we have

$\sigma=\$40,000$

sample size : n= 50

Then, the standard error of the mean will be :-

$SE=\dfrac{\$40000}{\sqrt{50}}=\$5656.85424949\approx\$5656.85$

Hence, the standard error of the mean= $5656.85

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Carla packed this box with 1 centimeter cubes what is the volume of the box

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Answer: $60\ cm^3$

Step-by-step explanation:

<h3> The complete exercise is attached.</h3>

You can observe in the picture attached that the box is a rectangular prism.

The volume of a rectangular prism can be found with this formula:

$V=lwh$

Where "l" is the length, "w" is the width and "h" is the height.

You know that the lenght of each side of those cubes is 1 centimeter. Therefore, you can multiply the number of cubes on each side of the box by 1 centimeter in order to find the lenght, the width and the height of the box:

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2 years ago

Which statements about the graph of the function y = (StartFraction 1 Over 3 EndFraction) Superscript x are true?

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the function is decreasing and the y intercept is (0,1)

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2 years ago

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2 years ago

A puppy finds a rawhide bone and begins to pull it with a force, Ft. The free-body diagram is shown. A free body diagram with 4

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Answer:

<em><u> It begins to move toward the right</u></em>

Explanation:

The given information can be summarized in this way:

First force vector: Fg = - 8N (vertical down)

Second force vector: Ft = 6N (horizonal right)

Third force vector: FN = 8N (vertical up)

Fourth vector: Ff = - 4N (horizontal left)

Following Newton's second law, net force equal mass times acceleration:

Net force = mass × acceleration

To predict the motion, you apply Newton's second law in each direction (vertical and horizontal)

<u>Vertical force balance:</u>

Net vertical force = 8N - 8N = 0. This means there is not motion in the vertical direction.

<u>Horizontal force balance:</u>

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1 year ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = <u><em>the length of the calls, in minutes.</em></u>

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = <u>0.3729</u>.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = <u>0.1271</u>

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = <u>0.1109</u>.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = <u>0.745</u>.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

2 years ago