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Mekhanik [1.2K]

3 years ago

11

Macy runs a 10k in 44 minutes. she plans to run a 15 k in a few months. how long will it take her to run a 15 k if she runs at t

he same pace

1 answer:

Nikolay [14]3 years ago

7 0

Answer:

$\frac{44 \times 15}{10} = 66 \: minutes$

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For the following exercise, determine the range (possible values) of the random variable, X. A batch of 300 machined parts conta

solmaris [256]

Answer:

{0,1,2,3,4,5}

Step-by-step explanation:

We are given that

Total number of machine parts=300

Number of defective machine parts=10

Total number of good machine parts=300-10=290

Sample contain parts that do not conform to customer requirement=5

X is a random variable which is the number of parts in a sample of 5 parts that do not conform to customer requirements.

We have to find the correct answer.

The sample contain 5 parts

Therefore, the possible values of random variable X

0,1,2,3,4,5

Hence, the range of X is given by

{0,1,2,3,4,5}

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3 years ago

Read the quotation from "Song of Myself."

maw [93]

Answer:

the answer is A

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the distance between california and mexico border is 100 miles less than thrice the distance between california and washington.

harina [27]

Answer: B. 3m-100

Step-by-step explanation:

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2 years ago

John answered 80% of the questions correctly. There were 30 questions on the test. How many questions did John answer correctly?

Liula [17]

Answer:

he got 24 right

Step-by-step explanation:

30 time 80% is 24

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3 years ago

The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g

Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

Step-by-step explanation:

This is a problem of the <em>distribution of sample means</em>. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves <em>normally</em> for samples sizes equal or greater than 30 $\\ n \geq 30$ . Mathematically

$\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a <em>normal standard distribution</em>, i.e., a normal distribution that has a population mean $\\ \mu = 0$ and a population standard deviation $\\ \sigma = 1$ .

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

From the question, we know that

The population mean is $\\ \mu = 88$ WPM
The population standard deviation is $\\ \sigma = 14$ WPM

We also know the size of the sample for this case: $\\ n = 137$ sixth graders.

We need to estimate the probability that a sample mean being greater than $\\ \overline{X} = 89.87$ WPM in the <em>distribution of sample means</em>. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}$

$\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}$

$\\ Z = 1.5634 \approx 1.56$

This is a <em>standardized value </em> and it tells us that the sample with mean 89.87 is 1.56<em> standard deviations</em> <em>above</em> the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any <em>cumulative</em> <em>standard normal table</em> available in Statistics books or on the Internet. Of course, this probability is the same that $\\ P(\overline{X} < 89.87)$ . Then

$\\ P(z$

However, we are looking for P(z>1.56), which is the <em>complement probability</em> of the previous probability. Therefore

$\\ P(z>1.56) = 1 - P(z$

$\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594$

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

5 0

3 years ago