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round to 4 decimal places as needed pls
This Question: 1 pt 20 of 24 ( complete This Test: 24 pts possible Suppose that an alline uses a seat with of 16.7 in. Assume
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Answer #1

Let X be the R.V that denotes the distribution of hip-breadth. Then X follows Norm(14.4,1)

Then by standardizing,

Practice Problems for Z-Scores

Then Z follows a standard normal distribution. Now by using the standard normal tables,

Mean

14.4
Standard Deviation 1
Probability for X > 16.7
X Value 16.7
Z Value 2.3
P(X>16.7) 0.0107

Hence P(X>16.7) = 0.0107

b)

Now we have a sample of 117 passengers and by CLT s,

\bar{X}\rightarrow N(\mu,\frac{\sigma}{\sqrt{n}})

And standardizing this, we have

Mean 14.4
Standard Deviation 0.09245
Probability for X_bar > 16.7
X Value 16.7
Z Value 24.878304
P(X_bar >16.7) ~0

Hence, P(X_bar > 16.7) is approximately zero.

c)

The result from B should be considered because only average individuals should be considered.

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