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2 years ago

5

At a large conference of teachers from a variety of subjects, a random sample of 50 mathematics teachers attending the conferenc

e was selected. Among the selected mathematics teachers, 28 percent had taken one or more courses in statistics. For which of the following populations is 28 percent a reasonable estimate of the percentage of those who have taken one or more courses in statistics?
A. All mathematics teachersB. All mathematics teachers who attended the conferenceC. All mathematics teachers who have taken one or more courses in statisticsD. All teachers who attended the conferenceE. All teachers

1 answer:

melamori03 [73]2 years ago

5 0

Answer:

C. All mathematics teachers who have taken one or more courses in statistics

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Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

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Answer:

Approximately, 159 men weighs more than 165 pounds and 159 men weighs less than 135 pounds.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 150 pounds

Standard Deviation, σ = 15

We are given that the distribution of weights of 1000 men is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P( men weighing more than 165 pounds)

P(x > 165)

$P( x > 165) = P( z > \displaystyle\frac{165 - 150}{15}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,

$P(x > 165) = 1 - 0.8413 = 0.1587 = 15.87\%$

Approximately, 159 men weighs more than 165 pounds.

P(men weighing less than 135 pounds)

P(x < 135)

$P( x < 135) = P( z < \displaystyle\frac{135 - 150}{15}) = P(z < -1)$

Calculation the value from standard normal z table, we have,

$P(x < 135) = 0.1587 = 15.87\%$

Approximately, 159 men weighs less than 135 pounds.

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2 years ago

Sarah was collecting money for charity. By Sunday night she had collected 55% of her target amount. On Monday she collected anot

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Answer = 350

Step-by-step explanation:

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Explain how you could write a quadratic function in factored form that would have a vertex with an x-coordinate of 3 and two dis

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Two distinct roots means two real solutions for x (the parabola needs to cross the x-axis twice)

Vertex form of a quadratic equation: (h,k) is vertex

y = a(x-h)^2 + k

The x of the vertex needs to equal 3

y = a(x-3)^2 + k

In order to have two distinct roots the parabola must be (+a) upward facing with vertex below the x-axis or (-a) downward facing with vertex above the x-axis. Parabolas are symmetrical so for an easy factorable equation make "a" 1 or -1 depending on if you want the upward/downward facing one.

y = (x-3)^2 - 1

Vertex (3,-1) upwards facing with two distinct roots 4 and 2

y = x^2 -6x + 9 - 1

y = x^2 -6x + 8

y = (x - 4)(x - 2)

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2 years ago

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A city’s bus line is used more as the urban population density increases. The more people in an area, the more likely bus lines

nevsk [136]

Answer:

(A) The residents of Belmont are more likely to use public transportation because the city has the highest population density.

Step-by-step explanation:

correct on edge

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2 years ago

A random sample from a normal population is obtained, and the data are given below. Find a 90% confidence interval for . 114 157

melamori03 [73]

Answer:

$103.160 \leq \sigma \leq 187.476$

And the upper bound rounded to the nearest integer would be 187.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

Data given: 114 157 203 257 284 299 305 344 378 410 421 450 478 480 512 533 545

The confidence interval for the population variance $\sigma^2$ is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^{17} (x_i -\bar x)^2}{n-1}}$

And in order to find the sample mean we just need to use this formula:

$\bar x =\frac{\sum_{i=1}^n x_i}{n}$

The sample mean obtained on this case is $\bar x= 362.941$ and the deviation s=132.250

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=17-1=16$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a tabel to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,16)" "=CHISQ.INV(0.95,16)". so for this case the critical values are:

$\chi^2_{\alpha/2}=26.296$

$\chi^2_{1- \alpha/2}=7.962$

And replacing into the formula for the interval we got:

$\frac{(16)(132.250)^2}{26.296} \leq \sigma \leq \frac{(16)(132.250)^2}{7.962}$

$10641.959 \leq \sigma^2 \leq 35147.074$

Now we just take square root on both sides of the interval and we got:

$103.160 \leq \sigma \leq 187.476$

And the upper bound rounded to the nearest integer would be 187.

4 0

2 years ago