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Phantasy [73]
3 years ago
7

Find Interquartile Range AND Range.

Mathematics
2 answers:
pav-90 [236]3 years ago
7 0

Answer:

IQR: 20

Range: 52

Step-by-step explanation:

IQR = 88 - 68 = 20

Range = 100 - 48 = 52

Alex_Xolod [135]3 years ago
5 0

(IQR) interquartile range: Upper quartile - Lower quartile = answer

Interquartile range: 88 - 68 = 20

Interquartile range = 20

Range: Highest value - Lowest value = answer

Range: 99 - 48 = 51

Range  = 51

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In the United States, 7% of all registered voters belong to the Green party. A random sample of 50 registered voters is taken. U
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Answer:

The expected value of the sample proportion is of 0.07.

1. P(p < .02) = 0.0823

2. P(p > .15) = 0.0132

3. P(.05 < p < .09) = 0.4176

Step-by-step explanation:

This question is solved using the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

7% of all registered voters belong to the Green party. 50 voters:

This means that p = 0.07, n = 50

So, for the normal distribution:

\mu = 0.07, s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.07*0.93}{50}} = 0.036

The expected value of the sample proportion is of 0.07.

1. Determine P(p < .02).

This is the pvalue of Z when X = 0.02. So

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

By the Central Limit Theorem

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

Z = \frac{0.02 - 0.07}{0.036}

Z = -1.39

Z = -1.39 has a pvalue of 0.0823

So

P(p < .02) = 0.0823

2. Determine P(p > .15).

This is 1 subtracted by the pvalue of Z when X = 0.15. So

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

Z = \frac{0.15 - 0.07}{0.036}

Z = 2.22

Z = 2.22 has a pvalue of 0.9868

1 - 0.9868 = 0.0132

So

P(p > .15) = 0.0132

3. Determine P(.05 < p < .09).

This is the pvalue of Z when X = 0.09 subtracted by the pvalue of Z when X = 0.05. So

X = 0.09

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

Z = \frac{0.09 - 0.07}{0.036}

Z = 0.55

Z = 0.55 has a pvalue of 0.7088

X = 0.05

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

Z = \frac{0.05 - 0.07}{0.036}

Z = -0.55

Z = -0.55 has a pvalue of 0.2912

0.7088 - 0.2912 = 0.4176. So

P(.05 < p < .09) = 0.4176

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