Jessica is 5 years older than her sister Jenna. Jenna tells her sister that in 5 years, she will be as old as Jessica was 5 year
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Answer:
(1). y = x ~ Exp (1/3).
(2). Check attachment.
(3). EY = 3(1 - e^-2).
(4). Var[y] = 3(1 - e^-2) (1 -3 (1 - e^-2)) - 36e^-2.
Step-by-step explanation:
Kindly check the attachment to aid in understanding the solution to the question.
So, from the question, we given the following parameters or information or data;
(A). The probability in which attempt to establish a video call via some social media app may fail with = 0.1.
(B). " If connection is established and if no connection failure occurs thereafter, then the duration of a typical video call in minutes is an exponential random variable X with E[X] = 3. "
(C). "due to an unfortunate bug in the app all calls are disconnected after 6 minutes. Let random variable Y denote the overall call duration (i.e., Y = 0 in case of failure to connect, Y = 6 when a call gets disconnected due to the bug, and Y = X otherwise.)."
(1). Hence, for FY(y) = y = x ~ Exp (1/3) for the condition that zero is equal to y = x < 6.
(2). Check attachment.
(3). EY = 3(1 - e^-2).
(4). Var[y] = 3(1 - e^-2) (1 -3 (1 - e^-2)) - 36e^-2.
The condition to follow in order to solve this question is that y = 0 if x ≤ 0, y = x if 0 ≤ x ≤ 6 and y = 6 if x ≥ 6.
Answer:
Step-by-step explanation:
Hello!
To test if boys are better in math classes than girls two random samples were taken:
Sample 1
X₁: score of a boy in calculus
n₁= 15
X[bar]₁= 82.3%
S₁= 5.6%
Sample 2
X₂: Score in the calculus of a girl
n₂= 12
X[bar]₂= 81.2%
S₂= 6.7%
To estimate per CI the difference between the mean percentage that boys obtained in calculus and the mean percentage that girls obtained in calculus, you need that both variables of interest come from normal populations.
To be able to use a pooled variance t-test you have to also assume that the population variances, although unknown, are equal.
Then you can calculate the interval as:
[(X[bar]_1-X[bar_2) ± *
]
[(82.3-81.2) ± 1.708* (6.11*]
[-2.94; 5.14]
Using a 90% confidence level you'd expect the interval [-2.94; 5.14] to contain the true value of the difference between the average percentage obtained in calculus by boys and the average percentage obtained in calculus by girls.
I hope this helps!