Consider n coins, each of which independently comes up heads with probability p. Suppose that n is large and p is small, and let λ = n p . Suppose that all n coins are tossed; if at least one comes up heads, the experiment ends; if not, we again toss all n coins, and so on. That is, we stop the first time that at least one of the n coins come up heads. Let X denote the total number of heads that appear. Which of the following reasonings concerned with approximating P { X = 1 } is correct (in all cases, y is a Poisson random variable with parameter λ ) a. Because the total number of heads that occur when all n coins are rolled is approximately a Poisson random variable with parameter P { X = 1 } ≈ P { Y = 1 } = λ e − λ b. Because the total number of heads that occur when all n coins are rolled is approximately a Poisson random variable with parameter λ , and because we stop only when this number is positive, P { X = 1 } ≈ P { Y = 1 | Y > 0 } = λ e − λ 1 − e − λ c. Because at least one coin comes up heads, X will equal I if none of the other n − 1 coins come up heads. Because the number of heads resulting from these n − 1 coins is approximately Poisson with mean ( n − 1 ) p ≈ λ , P { X = 1 } ≈ P { Y = 0 } = e − λ .
Consider n coins, each of which independently comes up heads with probability p. Suppose that n is large and p is small, and let λ = n p . Suppose that all n coins are tossed; if at least one comes up heads, the experiment ends; if not, we again toss all n coins, and so on. That is, we stop the first time that at least one of the n coins come up heads. Let X denote the total number of heads that appear. Which of the following reasonings concerned with approximating P { X = 1 } is correct (in all cases, y is a Poisson random variable with parameter λ ) a. Because the total number of heads that occur when all n coins are rolled is approximately a Poisson random variable with parameter P { X = 1 } ≈ P { Y = 1 } = λ e − λ b. Because the total number of heads that occur when all n coins are rolled is approximately a Poisson random variable with parameter λ , and because we stop only when this number is positive, P { X = 1 } ≈ P { Y = 1 | Y > 0 } = λ e − λ 1 − e − λ c. Because at least one coin comes up heads, X will equal I if none of the other n − 1 coins come up heads. Because the number of heads resulting from these n − 1 coins is approximately Poisson with mean ( n − 1 ) p ≈ λ , P { X = 1 } ≈ P { Y = 0 } = e − λ .
Solution Summary: The author explains that the option that is concerned with P(X=1) is in correct with the all cases.
Consider n coins, each of which independently comes up heads with probability p. Suppose that n is large and p is small, and let
λ
=
n
p
. Suppose that all n coins are tossed; if at least one comes up heads, the experiment ends; if not, we again toss all n coins, and so on. That is, we stop the first time that at least one of the n coins come up heads. Let X denote the total number of heads that appear. Which of the following reasonings concerned with approximating
P
{
X
=
1
}
is correct (in all cases, y is a Poisson random variable with parameter
λ
)
a. Because the total number of heads that occur when all n coins are rolled is approximately a Poisson random variable with parameter
P
{
X
=
1
}
≈
P
{
Y
=
1
}
=
λ
e
−
λ
b. Because the total number of heads that occur when all n coins are rolled is approximately a Poisson random variable with parameter
λ
, and because we stop only when this number is positive,
P
{
X
=
1
}
≈
P
{
Y
=
1
|
Y
>
0
}
=
λ
e
−
λ
1
−
e
−
λ
c. Because at least one coin comes up heads, X will equal I if none of the other
n
−
1
coins come up heads. Because the number of heads resulting from these
n
−
1
coins is approximately Poisson with mean
(
n
−
1
)
p
≈
λ
,
P
{
X
=
1
}
≈
P
{
Y
=
0
}
=
e
−
λ
.
Among a student group 54% use Google Chrome, 20% Internet Explorer, 10% Firefox, 5% Mozilla, and the rest use Safari. What is the probability that you need to pick 7 students to find 2 students using Google Chrome? Report answer to 3 decimals.
Samples of rejuvenated mitochondria are mutated (defective) with a probability 0.13. Find the probability that at most one sample is mutated in 10 samples. Report answer to 3 decimal places.
The same final exam of the astronomy course was given to two groups of students. The maximum number of points that a student can score is 100. The first group consisted of a random sample of 10 students who were taught by Professor A. Students from the first group obtained the following results:
87 88 91 88 86 92 81 93 73 99
The second group consisted of a random sample of 9 students who were taught by Professor B. Students from the second group obtained the following results:
74 74 79 97 67 88 86 83 78
Compute the mean squares of between-group variability, MSBET. Round your answer to two decimal places.
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