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For the binomial probability distribution with n = 5 and p = 0.1,

a) Find the probability that there are exactly 4 successes.
b) Find the probability that there are at least 4 successes.
c) Find the probability that there are at most 4 successes.
d) What is the shape of this binomial distribution?
A. Right-skewed
B. binompdf(5, 0.1, 4) = 0.00045
C. binomcdf(5, 0.1, 4) = 0.9999
D. 0.5
E. 1 - binomcdf(5, 0.1, 3) = 0.00046
F. 0.6708

e) What is the mean of this binomial distribution?
f) What is the standard deviation of this binomial distribution?

1 Answer

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Final answer:

The probability of exactly 4 successes in a binomial distribution with n=5 and p=0.1 is 0.00045. The probability of at least 4 successes is 0.00046, and the probability of at most 4 successes is 0.9999. The distribution is right-skewed, with a mean of 0.5 and a standard deviation of 0.67.

Step-by-step explanation:

To answer your questions regarding the binomial probability distribution with n = 5 and p = 0.1:

a) To find the probability of exactly 4 successes, we use the binomial probability formula P(X=k) = (n choose k) * p^k * (1-p)^(n-k), where X is the number of successes. Here, n = 5, p = 0.1, and k = 4. The calculation yields a probability of 0.00045.

b) The probability of at least 4 successes is the sum of the probabilities of exactly 4 and exactly 5 successes. This sum is equivalent to 1 minus the probability of 3 or fewer successes, which is 1 - binomcdf(5, 0.1, 3) = 0.00046.

c) The probability of at most 4 successes is binomcdf(5, 0.1, 4) = 0.9999.

d) The shape of this binomial distribution is right-skewed because the probability of success (0.1) is less than 0.5.

e) The mean of the binomial distribution is calculated with μ = np, which gives us 0.5.

f) The standard deviation is calculated with σ = √(npq), where q = 1 - p. The calculation yields a standard deviation of approximately 0.67.

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