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trapecia [35]
2 years ago
9

The heights of students in a class are normally distributed with mean 55 inches and standard deviation 5 inches. Use the Empiric

al Rule to determine the interval and contains the middle 68% of the heights.
a) [40,70]

b)[45,70]

c)[50,60]

d)[45,65]

e)[47,63]

d)none of the above
Mathematics
1 answer:
maks197457 [2]2 years ago
4 0

Answer:  c)[50,60]

Step-by-step explanation:

The Empirical rule says that , About 68% of the population lies with the one standard deviation from the mean (For normally distribution).

We are given that , The heights of students in a class are normally distributed with mean 55 inches and standard deviation 5 inches.

Then by Empirical rule, about 68% of the heights of students lies between one standard deviation from mean.

i.e. about 68% of the heights of students lies between \text{Mean}\pm\text{Standard deviation}

i.e. about 68% of the heights of students lies between 55\pm5

Here, 55\pm5=(55-5, 55+5)=(50,60)

i.e.  The required interval that contains the middle 68% of the heights. = [50,60]

Hence, the correct answer is c) (50,60)

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The U.S. Energy Information Administration (US EIA) reported that the average price for a gallon of regular gasoline is $2.94. T
Anit [1.1K]

Answer:

a) 25

b) 67

c) 97

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample. In this problem, \sigma = 0.25

(a) The desired margin of error is $0.10.

This is n when M = 0.1. So

M = z*\frac{\sigma}{\sqrt{n}}

0.1 = 1.96*\frac{0.25}{\sqrt{n}}

0.1\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.1}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.1})^{2}

n = 24.01

Rounding up to the nearest whole number, 25.

(b) The desired margin of error is $0.06.

This is n when M = 0.06. So

M = z*\frac{\sigma}{\sqrt{n}}

0.06 = 1.96*\frac{0.25}{\sqrt{n}}

0.06\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.06}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.06})^{2}

n = 66.7

Rounding up, 67

(c) The desired margin of error is $0.05.

This is n when M = 0.05. So

M = z*\frac{\sigma}{\sqrt{n}}

0.05 = 1.96*\frac{0.25}{\sqrt{n}}

0.05\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.05}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.05})^{2}

n = 96.04

Rounding up, 97

8 0
2 years ago
I would like to purchase 20 products at a cost of $65 per product. What would be my total with 3.5 sales tax
alexandr402 [8]

Answer:

Answer:

The total is: $ 1345.5

Step-by-step explanation:

It is given that:

I would like to purchase 20 products at a cost 65.00 per product.

This means that the cost of 20 products will be:

Also, there is a sales tax of 3.5%

This means that a person has to pay a extra 3.5% on the total cost of the items he purchased.

i.e. he need to pay 3/5% on $ 1300

This means that the amount of tax he need to pay is: 3.5% of 1300

                                                                             =  3.5%×1300

                                                                            = 0.035×1300

                                                                           = $ 45.5.

Hence, the total cost is: $ 1300+$ 45.5

This means that the total cost is: $ 134.5

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Meeting at least one person with the flu in thirteen random encounters on campus when the infection rate is 2% (2 in 100 people
s344n2d4d5 [400]

Answer:

23.1% probability of meeting at least one person with the flu

Step-by-step explanation:

For each encounter, there are only two possible outcomes. Either the person has the flu, or the person does not. The probability of a person having the flu is independent of any other person. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Infection rate of 2%

This means that p = 0.02

Thirteen random encounters

This means that n = 13

Probability of meeting at least one person with the flu

Either you meet none, or you meet at least one. The sum of the probabilities of these outcomes is 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). Then

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{13,0}.(0.02)^{0}.(0.98)^{13} = 0.7690

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.769 = 0.231

23.1% probability of meeting at least one person with the flu

6 0
2 years ago
Howard has a scale model of the Statue of Liberty. The model is 15 inches tall. The scale of the model to the actual statue is 1
sergiy2304 [10]

Answer:

Required equation \frac{1}{6.2}=\frac{15}{x}

The height of statue of liberty is 93 meters.

Step-by-step explanation:

Given : Howard has a scale model of the Statue of Liberty. The model is 15 inches tall. The scale of the model to the actual statue is 1 inch : 6.2 meters.

To find : Which equation can Howard use to determine x, the height in meters, of the Statue of Liberty?

Solution :

The model is 15 inches tall.

The scale of the model to the actual statue is 1 inch : 6.2 meters.

Let  x be the height in meters of the Statue of Liberty.

According to question, required equation is

\frac{1}{6.2}=\frac{15}{x}

Cross multiply,

x=15\times 6.2

x=93

Therefore, the height of statue of liberty is 93 meters.

4 0
2 years ago
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