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Rashid [163]
4 years ago
8

Suppose that you are conducting a survey on how many students have blue eyes in each first period classroom at Garfield Middle s

chool. The mean number of blue-eyed students is 6, and the standard deviation is 2. Mr. Garcia's first period history class has 10 students with blue eyes. What statement is true?
The number of blue-eyed students in Mr. Garcia's class is 2 standard deviations to the right of the mean,

The number of blue-eyed students in Mr. Garcia's class is 2 standard deviations to the left of the mean,

The number of blue-eyed students in Mr. Garcia's class is 4 standard deviations to the right of the mean,

The number of blue-eyed students in Mr. Garcia's class is 4 standard deviations to the left of the mean
Mathematics
1 answer:
rusak2 [61]4 years ago
4 0

Answer:

  • <u>The correct statement is the first one: </u><u><em>The number of blue-eyed students in Mr. Garcia's class is 2 standard deviations to the right of the mean</em></u><em> </em>

<em />

Explanation:

To calculate how many<em> standard deviations</em> a particular value in a group is from the mean, you can use the z-score:

      z-score=(x-\mu )/\sigma

Where:

  • z-score is the number of standard deviations the value of x is from the mean
  • \mu  is the mean
  • \sigma  is the standard deviation

Substitute in the formula:

       z-score=(10-6)/2=4/2=2

Which means that <em>the number of blue-eyed students in Mr. Garcia's class is 2 standard deviations</em> above the mean.

Above the mean is the same that to the right of the mean, because the in the normal standard probability graph the central value is Z = 0 (the z-score of the mean value is 0), the positive values are to the right of the central value, and the negative values are to the left of the central value.

Therefore, the correct statement is the first one: <em>The number of blue-eyed students in Mr. Garcia's class is 2 standard deviations to the right of the mean, </em>

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an Carlos County Health Department reported 17 new cases of tuberculosis in the first half of 2001 and 18 additional cases durin
Tpy6a [65]

Answer:

San Carlos County

Step-by-step explanation:

a) Data and Calculations:

Number of cases of tuberculosis reported in the first half of 2001 = 17

Number of cases of tuberculosis reported in the second half = 18

Total number of cases reported = 35 (17 + 18)

Population of the county at the beginning of 2001 = 204,500

Population of the county at the end of the year = 215,000

Average population for the year = 209,750 (204,500 + 215,000)/2

Therefore, the incidence rate of tuberculosis in the county = 35/209,750 * 100

= 0.000167

b) Since the incidence rate is <1, it suggests that the population of San Carlos County is at a very reduced risk of exposure to tuberculosis.  This means that only 1 in 6,000 (209,750/35) people are exposed to the disease in the county.

5 0
3 years ago
Solve the inequality if possible 4(3-2x)&gt;2(6-4x)
GenaCL600 [577]

Final Answer: No Solution

Steps/Reasons/Explanation:

Question: 4(3-  2x) > 2(6-  4x)

<u>Step 1</u>: Divide both sides by 2.

2(3 - 2x) > 6 - 4x

<u>Step 2</u>: Expand.

6 - 4x > 6 - 4x

<u>Step 3</u>: Since 6 - 4x > 6 - 4x is false, we have no solution.

No Solution

~I hope I helped you :)~

6 0
3 years ago
Read 2 more answers
Which of the following represents the amount a customer pays including the tip of 15% if the bill was b dollars? Select all that
masya89 [10]

Answer:

the third one!

Step-by-step explanation:

hope this helped!! :)

5 0
3 years ago
A very large batch of parts (assume a normal distrbution0 from a manufacturer has a mean weight = 43g and a standard deviation =
AVprozaik [17]

Answer:

5.44% probability that exactly 8 of the 16 parts you selected will have weights exceeding 45g

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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.

Percentage of parts with weights exceeding 45g?

1 subtracted by the pvalue of Z when X = 45. So

We have \mu = 43, \sigma = 4

Z = \frac{X - \mu}{\sigma}

Z = \frac{45 - 43}{4}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915

1 - 0.6915 = 0.3075

If you slect 16 parts at random form that batch, what is the probability that exactly 8 of the 16 parts you selected will have weights exceeding 45g?

This is P(X = 8) when n = 16, p = 0.3075. So

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

P(X = 8) = C_{16,8}.(0.3075)^{8}.(0.6915)^{8} = 0.0544

5.44% probability that exactly 8 of the 16 parts you selected will have weights exceeding 45g

4 0
3 years ago
Alex borrowed $12.50 from his friend Danilo. He paid him back $8.75. How much does he still owe? *
ollegr [7]

Answer:

$3.75

Step-by-step explanation: $12.50

                                             $  8.75

                                           ________

                                              $  3.75

8 0
3 years ago
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