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Lerok [7]
4 years ago
13

Assume that we want to construct a confidence interval. Do one of the​ following, as​ appropriate: (a) find the critical value t

Subscript alpha divided by 2​,​(b) find the critical value z Subscript alpha divided by 2​,or​ (c) state that neither the normal distribution nor the t distribution applies.Here are summary statistics for randomly selected weights of newborn​ girls: nequals235​,x overbarequals33.7​hg, sequals7.3hg. The confidence level is 95​%.
Mathematics
1 answer:
Svetradugi [14.3K]4 years ago
8 0

Answer:

To construct a confidence interval, Normal distribution should be used since the sample size is quite large (n > 30)

From the z-table, at α = 0.025 the critical value is

z_{\alpha/2} = 1.96

Step-by-step explanation:

We are given the following information:

The sample size is

n = 235

The mean weight is

\bar{x}= 33.7 \: hg

The standard deviation is

s = 7.3 \: hg

Since the sample size is quite large (n > 30) then according to the central limit theorem the sampling distribution of the sample mean will be approximately normal, therefore, we can use the Normal distribution for this problem.

The correct option is (b)

The critical value corresponding to 95% confidence level is given by

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

From the z-table, at α = 0.025 the critical value is

z_{\alpha/2} = 1.96

What is Normal Distribution?

A Normal Distribution is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.

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a 15 wire is tied from the top of a pole to a stake in the ground. if the stake is 5 ft from the base of the pole find the heigh
Ainat [17]

Imagine this pole of height h sticking out of the ground.  A wire of length 15 feet connects the top of the pole to a stake in the ground which is located 5 feet from the base of the pole.  This arrangement creates a right triangle with legs h (the height of the pole), 5 ft (distance of bottom of pole from stake) and hypotenuse 15 ft.  

We can find the height of the pole (h) using either trig or the Pythagorean Theorem.  If we use the P. T., then h^2 + 5^2 = 15^2.

This results in h^2 + 5^2 = 15^2, or h^2 + 25 = 225, and so:

h^2 = 200.  Thus, h = +√200 = +10√2.

The height of the pole is 10√2 ft, or approx. 14.14 ft.

5 0
3 years ago
Show all worked identify the asymptotes and state the end behavior of the function F(x)=6x over x-36
astraxan [27]

Solution

Asymptote:

Vertical Asymptote

- The vertical asymptotes of a rational function are determined by the denominator expression.

- The expression given is:

f(x)=\frac{6x}{x-36}

- The denominator of (x- 36) determines the asymptote line.

- The vertical asymptote defines where the rational function isundefined. Iin order for a rational function to be undefined, its denominator must be zero.

- Thus, we can say:

\begin{gathered} x-36=0 \\ Add\text{ 36 to both sides} \\  \\ \therefore x=36 \end{gathered}

- Thus, the vertical asymptote is

x=36

Horizontal Asymptote:

- The horizontal asymptote exists in two cases:

1. When the highest degree of the numerator is less han the degree of the demnominator. In this case, the horizontal asymptote is y = 0

2. When the highest degee sof the numerator and tdenominator are the same. In this case, the horizontal asymptote is

\begin{gathered} y=\frac{N}{D} \\ where, \\ N=\text{ Coefficient of the highest degree of the numerator} \\ D=\text{ Coefficient of the highest degree of the denominator} \end{gathered}

- For our question, we can see that the highest degrees of the numerator and denominator are the same. Thus, we have the Horizontal Asymptote to be:

y=\frac{6}{1}=6

End behavior:

- The end behavior is examining the y-values of the function as x tendsto negative and positive infinity.

- Thus, we have that:

\begin{gathered} f(x)=\frac{6x}{x-36} \\  \\ \text{ Divide top and bottom by }x \\ f(x)=\frac{6x}{x-36}\times\frac{x}{x} \\  \\ f(x)=\frac{\frac{6x}{x}}{\frac{x-36}{x}}=\frac{6}{1-\frac{36}{x}} \\  \\ As\text{ }x\to-\infty \\ f(-\infty)=\frac{6}{1-\frac{36}{-\infty}}=\frac{6}{1+\frac{36}{\infty}}=\frac{6}{1+0}=6 \\  \\ \text{ Thus, we can say: }x\to-\infty,f(x)\to6 \\  \\ Also, \\ As\text{  }x\to\infty \\ f(\infty)=\frac{6}{1-\frac{36}{\infty}}=\frac{6}{1-0}=6 \\  \\ \text{ Thus, we can also say: }x\to\infty,f(x)\to6 \end{gathered}

Final Answers

Asymptotes:

\begin{gathered} \text{ Vertical:} \\ x=36 \\  \\ \text{ Horizontal:} \\ y=6 \end{gathered}

End behavior:

\begin{gathered} As\text{  }x\to-\infty,f(x)\to6 \\  \\ As\text{  }x\to\infty,f(x)\to6 \end{gathered}

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