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ra1l [238]
3 years ago
14

Suppose that the IQs of university​ A's students can be described by a normal model with mean 150150 and standard deviation 77 p

oints. Also suppose that IQs of students from university B can be described by a normal model with mean 120120 and standard deviation 1010. ​a) Select a student at random from university A. Find the probability that the​ student's IQ is at least 140140 points.
Mathematics
1 answer:
NeX [460]3 years ago
3 0

Answer:

The probability that the​ student's IQ is at least 140 points is of 55.17%.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

University A: \mu = 150, \sigma = 77

a) Select a student at random from university A. Find the probability that the​ student's IQ is at least 140 points.

This is 1 subtracted by the pvalue of Z when X = 140. So

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

Z = \frac{140 - 150}{77}

Z = -0.13

Z = -0.13 has a pvalue of 0.4483.

1 - 0.4483 = 0.5517

The probability that the​ student's IQ is at least 140 points is of 55.17%.

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Answer:

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Step-by-step explanation:

The dependent <em>t</em>-test (also known as the paired <em>t</em>-test or paired samples <em>t</em>-test) compares the two means associated groups to conclude if there is a statistically significant difference amid these two means.

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The hypothesis for the test can be defined as follows:

<em>H₀</em>: With garlic​ treatment, the mean change in LDL cholesterol is not greater than 0, i.e. <em>d</em> ≤ 0.

<em>Hₐ</em>: With garlic​ treatment, the mean change in LDL cholesterol is greater than 0, i.e. <em>d</em> > 0.

The information provided is:

\bar d=0.40\\SD_{d}=16.2\\\alpha =0.01

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Decision rule:

If the <em>p</em>-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the <em>p</em>-value of the test as follows:

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*Use a <em>t</em>-table.

The <em>p</em>-value of the test is 0.4132.

<em>p-</em>value= 0.4132 > <em>α</em> = 0.01

The null hypothesis was failed to be rejected.

Thus, it can be concluded that with garlic​ treatment, the mean change in LDL cholesterol is not greater than 0.

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