Find the probability of this event Enter the answer as a fraction in simplest form, as a decimal, and as a
2 answers:
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Answer:
The point estimate for population mean is 8.129 Mpa.
Step-by-step explanation:
We are given the following in the question:
Data on flexural strength(MPa) for concrete beams of a certain type:
11.8, 7.7, 6.5, 6.8, 9.7, 6.8, 7.3, 7.9, 9.7, 8.7, 8.1, 8.5, 6.3, 7.0, 7.3, 7.4, 5.3, 9.0, 8.1, 11.3, 6.3, 7.2, 7.7, 7.8, 11.6, 10.7, 7.0
a) Point estimate of the mean value of strength for the conceptual population of all beams manufactured
We use the sample mean, as the point estimate for population mean.
Formula:
Thus, the point estimate for population mean is 8.129 Mpa.
Answer:
Length: 21 cm and Width: 6cm
Step-by-step explanation:
Let be the width and
length of the rectangle. We know:
-------- eq. 1
(Perimeter of the rectangle) ------- eq. 2
Substitute equation 1 in equation 2:
Using the value of y, we find x.
Answer:
Step-by-step explanation:
Hello!
For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.
In this example the variable is:
X: height of a college student. (cm)
There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.
The option you have is to apply the Central Limit Theorem.
The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:
X[bar]~~N(μ;σ2/n)
Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:
98% CI
1 - α: 0.98
⇒α: 0.02
α/2: 0.01
X[bar] ±
174.5 ±
[172.22; 176.78]
With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].
I hope it helps!