HERE IS YOUR ANS AFTER A LOT OF RESEARCH ! ^-^
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1 answer:
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Answer:
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- <u>Part A: No, you cannot.</u>
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- <u>Part B: 0.4491</u>
Explanation:
<u>Part A:</u>
The mean of samples of symmetrical (bell shaped) distributions follow a normal distribution pattern.
Thus, for symmetrical distributions you can use the z-score tables to make calculations that permit calculate probabilities for particular values.
For <em>skewed </em>distributions, in general, the samples do not follow a normal distribution pattern, except that the samples are large.
Since <em>a sample of 20 pictures</em> is not large enough, the answer to this question is negative: <em>you cannot accurately calculate the probability that the mean picture size is more than 3.8MB for an SRS (skewed right sample) of 20 pictures.</em>
<u>Part B.</u>
For large samples, the<em> Central Limit Theorem</em> will let you work with samples from skewed distributions.
Although the distribution of a population is skewed, the <em>Central Limit Theorem</em> states that large samples follow a normal distribution shape.
It is accepted that samples of 30 data is large enough to use the <em>Central Limit Theorem.</em>
Hence, you can use the z-core tables for standard normal distributions to calculate the probabilities for a <em>random sample of 60 pictures</em> instead of 20.
The z-score is calculated with the formula:
- z-score = (value - mean/ (standard deviation).
Here, mean = 3.7MB, value = 3.8MB, and standard deviation = 0.78MB.
Thus:
Then, you must use the z-score table to find the probability that the z-score is greater than - 0.128.
There are tables that show the cumulative probability in the right end and tables that show the cumulative probability in the left end of a normal standard distribution .
The probability that a z-score is greater than -0.128 is taken directly from a table with the cumulative probability in the left end. It is 0.4491.