A normal distribution is a type of continuous probability distribution for a real-valued random variable in statistics.
Yes, the large-sample confidence interval will be valid.
<h3>What is meant by normal distribution?</h3>
A normal distribution is a type of continuous probability distribution for a real-valued random variable in statistics.
The normal distribution, also known as the Gaussian distribution, is a symmetric probability distribution about the mean, indicating that data near the mean occur more frequently than data far from the mean.
The confidence interval will be valid regardless of the shape of the population distribution as long as the sample is large enough to satisfy the central limit theorem.
<h3>
What does a large sample confidence interval for a population mean?</h3>
A sample is considered large when n ≥ 30.
By 'valid', it means that the confidence interval procedure has a 95% chance of producing an interval that contains the population parameter.
To learn more about normal distribution, refer to:
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Answer:
Part 1:
or 
Part 2: 
Step-by-step explanation:
To start things off, you must put everything on the opposite side of y so that you only have the y variable left.
In this case, put x onto the other side, making it negative since it's flipped, so you get
(since the x is negative you're using subtraction from 11).
Next, divide both sides by 5 to make 5y into y:


Then in this case you turn y=f(x), but the system did it for you, so all you have to put is
.
For the second part you must replace x with 2 since you need to find f(2).

Your answer for part 2 is
.
Answer:
In order to find the variance we need to calculate first the second moment given by:
And the variance is given by:
![Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20%2B%5BE%28X%29%5D%5E2%20%3D%2023.36%20-%5B4.74%5D%5E2%20%3D%200.8924)
And the deviation would be:

Step-by-step explanation:
Previous concepts
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
Solution to the problem
For this case we have the following distribution given:
X 3 4 5 6
P(X) 0.07 0.4 0.25 0.28
We can calculate the mean with the following formula:

In order to find the variance we need to calculate first the second moment given by:

And the variance is given by:
![Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20%2B%5BE%28X%29%5D%5E2%20%3D%2023.36%20-%5B4.74%5D%5E2%20%3D%200.8924)
And the deviation would be:

Answer:
c
Step-by-step explanation: