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pav-90 [236]
3 years ago
6

An elementary school class ran 1 mile in an average of 11 minutes with a standard deviation of 3 minutes. Rachel, a student in t

he class, ran 1 mile in 8 minutes. A junior high school class ran 1 mile in an average of 9 minutes, with a standard deviation of 2 minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes. A high school class ran 1 mile in an average of 7 minutes with a standard deviation of 4 minutes. Nedda, a student in the class, ran 1 mile in 8 minutes. Who is the fastest runner with respect to his or her class?
Mathematics
1 answer:
EleoNora [17]3 years ago
8 0

Answer:

Rachel

Step-by-step explanation:

We need to measure how far (towards the left) are the students from the mean in<em> “standard deviations units”</em>.  

That is to say, if t is the time the student ran the mile and s is the standard deviation of the class, we must find an x such that

mean - x*s = t

For Rachel we have

11 - x*3 = 8, so x = 1.  

Rachel is <em>1 standard deviation far (to the left) from the mean</em> of her class

For Kenji we have

9 - x*2 = 8.5, so x = 0.25

Kenji is <em>0.25 standard deviations far (to the left) from the mean</em> of his class

For Nedda we have

7 - x*4 = 8, so x = 0.25

Nedda is also 0.25 standard deviations far (to the left) from the mean of his class.

As Rachel is the farthest from the mean of her class in term of standard deviations, Rachel is the fastest runner with respect to her class.

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What is the slope of points (4,9) (-8,-6)? ​
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Answer:

5/4

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(y₂ - y₁) / (x₂ - x₁)

(4,9) (-8, -6)

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A national survey of companies included a question that asked whether the company had at least one bilingual telephone operator.
nirvana33 [79]

Answer:

The first option is correct. Option A is correct.

LCL = 0.270, and UCL = 0.397

80% Confidence interval = (0.270, 0.397)

Step-by-step explanation:

The data for Y and N for the 90 companies is attached to this solution provided.

Y represents companies with at least 1 bilingual operator and N represents companies with no bilingual operator.

The number of Y in the data = 30

Hence, sample proportion of companies with at least one bilingual operator = (30/90) = 0.3333

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

Mathematically,

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

Sample proportion = 0.3333

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)

Critical value at 80% confidence level for sample size of 90 is obtained from the z-tables.

Critical value = 1.280

Standard error of the mean = σₓ = √[p(1-p)/n]

p = sample proportion

n = sample size = 90

σₓ = √[0.3333×0.6667)/90] = 0.0496891568 = 0.04969

80% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.3333 ± (1.28 × 0.04969)

CI = 0.3333 ± 0.0636021207

80% CI = (0.2696978793, 0.3969021207)

80% Confidence interval = (0.270, 0.397)

Hope this Helps!!!

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