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Sagot :
Answer:
0.2061 = 20.61% probability, given a class size of 30 students, the average time to complete the test is less than 48.5 minutes
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean of 50 minutes and a standard deviation of 10 minutes.
This means that [tex]\mu = 50, \sigma = 10[/tex]
Class size of 30 students
This means that [tex]n = 30, s = \frac{10}{\sqrt{30}}[/tex]
What is the probability, given a class size of 30 students, the average time to complete the test is less than 48.5 minutes.
This is the p-value of Z when X = 48.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{48.5 - 50}{\frac{10}{\sqrt{30}}}[/tex]
[tex]Z = -0.82[/tex]
[tex]Z = -0.82[/tex] has a p-value of 0.2061
0.2061 = 20.61% probability, given a class size of 30 students, the average time to complete the test is less than 48.5 minutes
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