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tiny-mole [99]
2 years ago
6

The conditional relative frequency table was calculated by row using data from a survey of one station’s television programmin

g. The survey compared the target audience with the type of show, either live or recorded, over a 24-hour time period. A 4-column table with 3 rows titled television programming. The first column has no label with entries adult, children, total. The second column is labeled live with entries 0. 625, 0. 5, nearly equal to 0. 583. The third column is labeled recorded with entries 0. 375, 0. 5, nearly equal to 0. 417. The fourth column is labeled total with entries 1. 0, 1. 0, 1. 0. Which is the best description of the value 0. 375 in the table? Given that the program was targeted at adults, there is a 37. 5% chance that it was recorded. Given that the program was recorded, there is a 37. 5% chance that it was targeted at adults. 37. 5% of the programs are targeted at adults. 37. 5% of the programs are recorded.
Mathematics
1 answer:
Setler [38]2 years ago
5 0

The conditional relative frequency presents the result of the survey that conveys information about the programming and the audience.

  • The best description of the value 0.357 in the table is; <u>Given that the program was targeted at adults, there is a 37.5% chance that it was recorded</u>.

Reasons:

The table is presented as follows;

\begin{tabular}{|c|c|c|c|}&Live&Recorded&Total\\Adult&0.625&0.375&1.0\\Children&0.5&0.5&1.0\\Total&\approx 0.583&\approx 0.417&1.0\end{array}\right]

Based on the given table, we have that the programs are categorized based on the intended audience.

Therefore, the 0.375, indicates that 37.5% of the program targeted at adults are recorded.

Therefore, the correct option is;

<u>Given that the program was targeted at adults, there is a 37.5% chance that it was recorded.</u>

Learn more about relative frequency table here:

brainly.com/question/8470699

brainly.com/question/3712144

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Answer:

a) Nothing, beause the distribution of the monthly rental prices are not normal.

b) 1.43% probability that the sample mean rent price will be greater than $900

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:

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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 random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

(a) Suppose a one-bedroom rental listing in this large city is selected at random. What can be said about the probability that the listed rent price will be at least $930?

Nothing, beause the distribution of the monthly rental prices are not normal.

(b) Suppose a random sample 30 one-bedroom rental listing in this large city will be selected, the rent price will be recorded for each listing, and the sample mean rent price will be computed. What can be said about the probability that the sample mean rent price will be greater than $900?

Now we can apply the Central Limit Theorem.

\mu = 880, \sigma = 50, n = 30, s = \frac{50}{\sqrt{30}} = 9.1287

This probability is 1 subtracted by the pvalue of Z when X = 900.

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

By the Central Limit Theorem

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

Z = \frac{900 - 880}{9.1287}

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Z = 2.19 has a pvalue of 0.9857

1 - 0.9857 = 0.0143

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