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Tasya [4]
2 years ago
15

The amount of Jen’s monthly phone bill is normally distributed with a mean of $75 and a standard deviation of $8. What percentag

e of her phone bills are between $51 and $99?
A. 99.7%
B. 95%
C. 99.99%
D. 68%
Mathematics
1 answer:
nikklg [1K]2 years ago
8 0
Answer:a is the answer
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How does using a bar diagram help you in predicting the solution to ratio and rate problems?
jeka94
Bar graph display directly the varibales which are the rate and ratio of the numbers to visualize and display the results, these contrasting the different outcomes. On the contrary, histogram is used in grouped frequency parameters. Moreover, as little as the given five parameters or data set this will be ineffective and will result to a bar graph only and basically, the suited option is the aforementioned vertical graph to display the numbers. To expound on the definition of histogram it is used when the frequency is grouped. For example the data set of 1-5, 6-10, 11-15 and 16-20 this now can be used and applied to illustrate histogram because of the number and quantity of the given data.<span>
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3 0
3 years ago
A random sample of 20 recent weddings in a country yielded a mean wedding cost of $ 26,388.67. Assume that recent wedding costs
Makovka662 [10]

Answer:

a) 95% confidence interval for the mean​cost, μ​, of all recent weddings in this country = (22,550.95, 30,226.40)

.The​ 95% confidence interval is from $22,550.95 to $30,226.40.

b) For the interpretation of the result, option D is correct.

We can be​ 95% confident that the mean​ cost, μ​, of all recent weddings in this country is somewhere within the confidence interval.

c) Option B is correct.

The population mean may or may not lie in this​ interval, but we can be​ 95% confident that it does.

Step-by-step explanation:

Sample size = 20

Sample Mean = $26,388.67

Sample Standard deviation = $8200

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = 26,388.67

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 20 - 1 = 19.

Significance level for 95% confidence interval

(100% - 95%)/2 = 2.5% = 0.025

t (0.025, 19) = 2.086 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 8200

n = sample size = 20

σₓ = (8200/√20) = 1833.6

99% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 26,388.67 ± (2.093 × 1833.6)

CI = 26,388.67 ± 3,837.7248

99% CI = (22,550.9452, 30,226.3948)

99% Confidence interval = (22,550.95, 30,226.40)

a) 95% confidence interval for the mean​cost, μ​, of all recent weddings in this country = (22,550.95, 30,226.40)

.The​ 95% confidence interval is from $22,550.95 to $30,226.40.

b) The interpretation of the confidence interval obtained, just as explained above is that we can be​ 95% confident that the mean​ cost, μ​,of all recent weddings in this country is somewhere within the confidence interval

c) A further explanation would be that the population mean may or may not lie in this​ interval, but we can be​ 95% confident that it does.

Hope this Helps!!!

4 0
2 years ago
A student leaving school walks 2.8 km north and then walks 1.2 km south. What is
vova2212 [387]

Answer: 1.6km

Step-by-step explanation: 2.8km - 1.2km

The student is still north from school.

I hope this answer helped! :)

4 0
3 years ago
Read 2 more answers
Sandi buys 2 fewer packages of pencils than p packages of pens. Pencils costs $2.45 per package and pens costs $3 per package. W
Ilya [14]
I think its B Because u have the total of pencils and how much they cost now u need to how know many pins you have to know how many they have in total
3 0
3 years ago
A survey showed that 77% of us need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. 13 adults are randomly
earnstyle [38]

Using the binomial distribution, it is found that the probability that at least 12 of the 13 adults require eyesight correction is of 0.163 = 16.3%. Since this probability is greater than 5%, it is found that 12 is not a significantly high number of adults requiring eyesight correction.

For each person, there are only two possible outcomes, either they need correction for their eyesight, or they do not. The probability of a person needing correction is independent of any other person, hence, the binomial distribution is used to solve this question.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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

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

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • A survey showed that 77% of us need correction, hence p = 0.77.
  • 13 adults are randomly selected, hence n = 13.

The probability that at least 12 of them need correction for their eyesight is given by:

P(X \geq 12) = P(X = 12) + P(X = 13)

In which:

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

P(X = 12) = C_{13,12}.(0.77)^{12}.(0.23)^{1} = 0.1299

P(X = 13) = C_{13,13}.(0.77)^{13}.(0.23)^{0} = 0.0334

Then:

P(X \geq 12) = P(X = 12) + P(X = 13) = 0.1299 + 0.0334 = 0.163

The probability that at least 12 of the 13 adults require eyesight correction is of 0.163 = 16.3%. Since this probability is greater than 5%, it is found that 12 is not a significantly high number of adults requiring eyesight correction.

More can be learned about the binomial distribution at brainly.com/question/24863377

7 0
2 years ago
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