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Gnom [1K]
3 years ago
11

The 20 tallest buildings in the world have an average height of 1,667.25 feet, with a standard deviation of 318.31 feet. The Zif

eng Tower and the Wuhan Center have heights of 1,476 feet and 1,437 feet, respectively. The Shanghai Tower and Marina 101 have z-scores of 1.27 and −0.84, respectively.
Part A: List these four buildings in order from shortest to tallest.


Part B: Mathematically explain your reasoning and be sure to include all necessary calculations.
Mathematics
1 answer:
valentina_108 [34]3 years ago
6 0

Answer:

Marina, Wuhan center, Zifeng Tower , Shangai Tower

Step-by-step explanation:

Using the z-scores of the Shangai tower and the Marina, we can calculate their heights

Mathematically, the z-score or standard score can be calculated as follows

Zscore =  X - μ/σ

where σ and μ represents the standard deviation and the population mean respectively.

For Shaghai Tower ;

1.27 = (X - 1667.25)/318.31

318.31 × 1.27 = X - 1667.25

404.25 = X - 1667.25

X = 1667.25 + 404.25 = 2071.50

The height of the Shaghai Tower is 2071.50 feet

For Marina, we use the same procedure;

-0.84 = (X - 1667.25)/318.31

318.31 × -0.84 = X - 1667.25

-267.38 = X - 1667.25

X = 1667.25 - 267.38 = 1399.87

The height of Marina building is 1399.87

Listing the buildings' height in ascending order we have; Marina, Wuhan center, Zifeng Tower , Shangai Tower

The reasoning behind this can be mathematically explained in terms of the z-score. The standard score of the heights of the building give us an insight as to how tall the buildings can be.

Since they are in the same population i.e building heights, the higher the z-score, the higher the height of the building. A positive z-score indicates a height taller than the average while a negative z-score indicate a height lower than the average height which is the mean

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Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

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The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305

The degrees of freedom for this sample size are:

df=n-1=50-1=49

The t-value for a 95% confidence interval and 49 degrees of freedom is t=2.01.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.01 \cdot 0.305=0.61

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 6.34-0.61=5.73\\\\UL=M+t \cdot s_M = 6.34+0.61=6.95

The 95% confidence interval for the mean is (5.73, 6.95).

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