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Sagot :
Answer:
a) P(X < 99) = 0.2033.
b) P(98 < X < 100) = 0.4525
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 estabilishes 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 100 and variance of 36.
This means that [tex]\mu = 100, \sigma = \sqrt{36} = 6[/tex]
Sample of 25:
This means that [tex]n = 25, s = \frac{6}{\sqrt{25}} = 1.2[/tex]
(a) P(X<99)
This is the pvalue of Z when X = 99. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{99 - 100}{1.2}[/tex]
[tex]Z = -0.83[/tex]
[tex]Z = -0.83[/tex] has a pvalue of 0.2033. So
P(X < 99) = 0.2033.
b) P(98 < X < 100)
This is the pvalue of Z when X = 100 subtracted by the pvalue of Z when X = 98. So
X = 100
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{100 - 100}{1.2}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] has a pvalue of 0.5
X = 98
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{98 - 100}{1.2}[/tex]
[tex]Z = -1.67[/tex]
[tex]Z = -1.67[/tex] has a pvalue of 0.0475
0.5 - 0.0475 = 0.4525
So
P(98 < X < 100) = 0.4525
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