Question

Suppose that $X_1$, $X_2$, $X_3$ and $X_4$ are independent normal variables such that $X_1 \sim N(1, 2)$, $X_2 \sim N(2, 3)$, $X_3 \sim N(3, 6)$ and $X_4 \sim N(4, 5)$. a. Find the probability of the event \"$X_1$ lies between 0 and 1 and $X_2$ lies between 1 and 2\". b. Find the probability that the sum of $X_1$ and $X_2$ is less than $X_3$. c. Let $Y_1$ be the average of $X_1$ and $X_2$, and $Y_2$ be the average of $X_3$ and $X_4$. What is the probability that $Y_1$ and $Y_2$ will differ by at least 1, i.e. that $P(|Y_1 - Y_2| \ge 1)$? d. Find $a$ such that $P(\sum_{i=1}^{4} X_i \le a) = 0.9$.

          Suppose that $X_1$, $X_2$, $X_3$ and $X_4$ are independent normal variables such that $X_1 \sim N(1, 2)$, $X_2 \sim N(2, 3)$, $X_3 \sim N(3, 6)$ and $X_4 \sim N(4, 5)$.

a. Find the probability of the event \"$X_1$ lies between 0 and 1 and $X_2$ lies between 1 and 2\".
b. Find the probability that the sum of $X_1$ and $X_2$ is less than $X_3$.
c. Let $Y_1$ be the average of $X_1$ and $X_2$, and $Y_2$ be the average of $X_3$ and $X_4$. What is the probability that $Y_1$ and $Y_2$ will differ by at least 1, i.e. that $P(|Y_1 - Y_2| \ge 1)$?
d. Find $a$ such that $P(\sum_{i=1}^{4} X_i \le a) = 0.9$.

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Suppose that X1, X2, X3 and X4 are independent normal variables such that X1 ∼ N(1, 2), X2 ∼ N(2, 3), X3 ∼ N(3, 6) and X4 ∼ N(4, 5).

a. Find the probability of the event $̈X1lies between 0 and 1 andX2lies between 1 and 2.̈
b. Find the probability that the sum ofX1andX2is less thanX3.
c. LetY1be the average ofX1andX2, andY2be the average ofX3andX4. What is the probability thatY1andY2will differ by at least 1, i.e. thatP(|Y1 - Y2| ≥1)?
d. Findasuch thatP(∑i=1^4 Xi ≤a) = 0.9.

Added by Rocio S.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Supreeta N

11/06/2023

Step 1

" To find this probability, we need to calculate the individual probabilities for X1 and X2 and then multiply them together since X1 and X2 are independent. P(X1 lies between 0 and 1) = P(0 < X1 < 1) = P(X1 < 1) - P(X1 < 0) Using the standard normal distribution Show more…

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Please answer the following question. Show all steps and calculations. Suppose that X1, X2, X3, and X4 are independent normal variables such that X1 ~ N(1,2), X2 ~ N(2,3), X3 ~ N(3,6), and X4 ~ N(4,5). a. Find the probability of the event "X1 lies between 0 and 1 and X2 lies between 1 and 2." b. Find the probability that the sum of X1 and X2 is less than X3. c. Let Y be the average of X1 and X2, and Z be the average of X3 and X4. What is the probability that Y and Z will differ by at least 1, i.e. that P(Y - Z > 1)? d. Find a such that P(X < a) = 0.9.

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