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3 years ago

Please help me on this!

Mathematics

1 answer:

trapecia [35]3 years ago

8 0

Answer:

1) Jina is not a member of the Music Club.

2) Three students are a member of the Music Club but not the Math Club.

3 Mary and Dan are in all three clubs.

Step-by-step explanation:

Look at the diagram and notice how each club has its own circle. If a student is in that circle, then they are a member of that club. When circles overlap, that overlap space means that a student is a member of two or all three clubs, depending on how many and which circles overlap.

1) Looking at Jina's location on the diagram, she is in the overlap between the math and chess club circles. However, her position does NOT overlap with the Music Club circle, therefore she is not a member of the Music Club.

2) Let's look at the students of the Music Club circle and its overlaps with the Chess Club circle, but NOT the overlaps with the Math Club circle. We can see that Lucy, Josh, and Juan are in these locations, therefore there are three students who are members of the Music Club but not the Math Club.

3) To find which students are in all three clubs, you would look at the middle of the diagram where all three circles overlap. The students listed in this location are Mary and Dan, therefore they are in all three clubs.

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kotegsom [21]

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Step-by-step explanation:

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3 years ago

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tangare [24]

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Since you find the area by multiplying things together, for example x · x = x², the exponent is there.

You do not need to worry about this being in the calculation. This will just be in units such as:

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7 0

3 years ago

Read 2 more answers

The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation

fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 197.5, \sigma = 8.3$

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

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

$Z = \frac{210.9 - 197.5}{8.3}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now $n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14$

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{196 - 197.5}{2.14}$

$Z = -0.7$

$Z = -0.7$ has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0

3 years ago