Answer:
Step-by-step explanation:
We are given the following information in the question.
The open interval: (- 18,22).
Let p belong to the given interval then p represents all the values belonging to the given interval.
Open Interval: An open interval is a interval that does not include the end points.
Thus, we can write:
is the required inequality form required.
That is p can take values greater than -18 and smaller than 22.
Answer:
-1 + 4
Step-by-step explanation:
i did the math
Answer: The probability in (b) has higher probability than the probability in (a).
Explanation:
Since we're computing for the probability of the sample mean, we consider the z-score and the standard deviation of the sampling distribution. Recall that the standard deviation of the sampling distribution approximately the quotient of the population standard deviation and the square root of the sample size.
So, if the sample size higher, the standard deviation of the sampling distribution is lower. Since the sample size in (b) is higher, the standard deviation of the sampling distribution in (b) is lower.
Moreover, since the mean of the sampling distribution is the same as the population mean, the lower the standard deviation, the wider the range of z-scores. Because the standard deviation in (b) is lower, it has a wider range of z-scores.
Note that in a normal distribution, if the probability has wider range of z-scores, it has a higher probability. Therefore, the probability in (b) has higher probability than the probability in (a) because it has wider range of z-scores than the probability in (a).
Looks like a badly encoded/decoded symbol. It's supposed to be a minus sign, so you're asked to find the expectation of 2<em>X </em>² - <em>Y</em>.
If you don't know how <em>X</em> or <em>Y</em> are distributed, but you know E[<em>X</em> ²] and E[<em>Y</em>], then it's as simple as distributing the expectation over the sum:
E[2<em>X </em>² - <em>Y</em>] = 2 E[<em>X </em>²] - E[<em>Y</em>]
Or, if you're given the expectation and variance of <em>X</em>, you have
Var[<em>X</em>] = E[<em>X</em> ²] - E[<em>X</em>]²
→ E[2<em>X </em>² - <em>Y</em>] = 2 (Var[<em>X</em>] + E[<em>X</em>]²) - E[<em>Y</em>]
Otherwise, you may be given the density function, or joint density, in which case you can determine the expectations by computing an integral or sum.
