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kotykmax [81]
3 years ago
8

As part of a class project, a university student surveyed the students in the cafeteria lunch line to look for a relationship be

tween eye color and hair color among students. The table below contains the results of the survey.
Eye Color
Hair
Color Blue Gray Green Brown Marginal
Totals
Blond 42 5 21 10 78
Red 12 22 19 12 65
Brown 22 5 12 34 73
Black 9 3 11 64 87
Marginal
Totals 85 35 63 120 303

From the sample population of students with gray eyes, what is the relative frequency of students with red hair?


0.07


0.27


0.33


0.26
Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
6 1

Answer:

0.63

Step-by-step explanation:

To get the relative frequency of students with red hair, we need to find the relative frequency of students with red hair from the students with gray eyes.

Since we already know that the total number of students with gray eyes is 35 and the number of red hair students with gray eye is 22

The formula we will use to calculate the relative frequency will be

The relative frequency of students with red hair = Total students of red hair with gray eyes divide by total number of gray eye students

22 ÷ 35 = 0.628  ≅ 0.63

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At a new exhibit in the Museum of Science, people are asked to choose between 94 or 220 random draws from a machine. The machine
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Answer:

0.0869 = 8.69% probability of getting more than 61% green balls.

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, 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.

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}}

The machine is known to have 99 green balls and 78 red balls.

This means that p = \frac{99}{99+78} = 0.5593

Mean and standard deviation:

\mu = p = 0.5593

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5593*0.4407}{99+78}} = 0.0373

a. Calculate the probability of getting more than 61% green balls.

This is 1 subtracted by the pvalue of Z when X = 0.61. So

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

By the Central Limit Theorem

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

Z = \frac{0.61 - 0.5593}{0.0373}

Z = 1.36

Z = 1.36 has a pvalue of 0.9131

1 - 0.9131 = 0.0869

0.0869 = 8.69% probability of getting more than 61% green balls.

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A pair of $36 jeans were on sale at Jim's Jeans for 1/3 off. The same pair of jeans were $42 at David's Denims. The jeans were o
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Complete question :

A pair of $36 jeans were on sale at Jim's Jeans for 1/3 off. The same pair of jeans were $42 at David's Denims. The jeans were on sale for 2/3 off.

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