Question

For a given binomial distribution with n = 11, p = 0.35, find the following probabilities P(x is greater than 7, x> 7)

Answer #1

)Solution

Given that ,

p = 0.35

q = 1 - p = 1 - 0.35 = 0.65

n = 11

Using binomial probability formula ,

P(X = x) = (n C x) * p x * (1 - p)n - x

P(X > 7) = 1 - P(X 7)

= 1 - P(X = 0) - P(X = 1)- P(X = 2)- P(X = 3)- P(X = 4)- P(X = 5)- P(X = 6)- P(X = 7)

= 1 - (11 C 0) * 0.35 0 * (0.65)11 - (11 C 1) * 0.35 1 * (0.65)10 - (11 C 2) * 0.35 2 * (0.65)9 - (11 C 3) * 0.35 3 * (0.65)8 - (11 C 4) * 0.35 4 * (0.65)7 - (11 C 5) * 0.35 5 * (0.65)6 - (11 C 6) * 0.35 6 * (0.65)4 - (11 C 7) * 0.35 7 * (0.65)5

= 1 - 0.9878

= 0.0122

Probability = 0.0122

For a given binomial distribution with n =
11, p = 0.35, find the following probabilities P(
x is less than or equal to 6, x≤ 6)

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(Round your answers to 4 decimal places.)
A) x = 2
B) x = 3

In a binomial distribution, n = 7 and π=0.24π=0.24 .
Find the probabilities of the following events. (Round your
answers to 4 decimal places.)
a.
x=2x
b.
x≤2x
c.
x≥3x

In a binomial distribution, n=7 and π=.15. Find the
probabilities of the following events. (Round your answers to 4
decimal places.)
(a) x=1
Probability=
(b) x≤2
Probability =
(c) x≥3
Probability=

A random variable follows a binomial distribution. Given n = 4
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Using the Binomial distribution,
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distribution and geometric distribution. Group of answer choices
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0.08 Binomial distribution: P(3) = 0.181. Geometric distribution:
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approximate the following binomial probabilities for
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p(x=18,n=50,p=0.3)
p(x>15,n=50,p=0.3)
p(x>12,n=50,p=0.3)
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1. Show that the normal approximation to the binomial can
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Make continuity corrections for each of the
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P(x ≤ 95)
P(x < 65)
P(x ≥ 100)
P(x > 100)

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n=6. Determine the probabilities below. Round to four decimal
places as needeed.
a) P(x=2)
b) P(x< or equal to1)
c) P(x>4)

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