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laiz [17]
2 years ago
6

A given set of values is found to be a normal distribution with a mean of 140 and a standard deviation of 18.0. Find the value t

hat is greater than 45% of the data values.
Mathematics
2 answers:
Andreyy892 years ago
6 0

Answer:

Formula for Z score

Z=\frac{X-B}{A}

Where, Z is Z score for value that is greater than 45% of the data values.

X=Score=?

B=Mean =140

Standard Deviation = 18

Z score for value above 45 % of data set = 0.9987 - 0.0668=0.9319

Z_{45 percent above}=Z_{100}-Z_{45}=0.9987-0.0668=0.9319\\\\0.9319=\frac{X-140}{18}\\\\ 0.9319*18=X-140\\\\X=140 +16.7742\\\\ X=156.7742

X score for value that is greater than 45% of the data values.= 156.78 (Approx)

Alisiya [41]2 years ago
5 0

Answer:

The value that is greater than 45% of the data values is approximately 137.84.

Step-by-step explanation:

The key is transforming values from this distribution to a z-score range and finding the corresponding value using a z-score table.

We are looking for a value x which attains a critical z-score that corresponds to the (100-45)%=55-th percentile:

z_{0.55} = \frac{x-\mu}{\sigma}=\frac{x-140}{18}\implies x = 18\cdot z_{0.55}+140

The critical z value (from z-score table, online) is: -0.12, so:

x = 18\cdot z_{0.55}+140=18\cdot(-0.12)+140\approx137.84

The value that is greater than 45% of the data values is approximately 137.84.


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I have attached the graph of each of the function, you can look at it for visualization.

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Option B: \frac{1}{\sqrt[3]{x^{5} } }

The expression \frac{1}{\sqrt[3]{x^{5} } } is equivalent to the simplified expression  \frac{1}{\sqrt[3]{x^{5} } }

Thus, the expression \frac{1}{\sqrt[3]{x^{5} } } is equivalent to x^{-\frac{5}{3}

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Thus, the expression -\sqrt[3]{x^5} is not equivalent to x^{-\frac{5}{3}

Hence, Option C is not the correct answer.

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Thus, the expression -\sqrt[5]{x^3} is not equivalent to x^{-\frac{5}{3}

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