Question

In a hamster breeders experience. the number of live pups in a litter of a famale not over twelve months in age who had not borne a litter I the past six week

has the following probabilities distribution.

X 3 4 5 6 7 8 9
p(x) 0.04 0.10 0.26 0.31 0.22 0.05 0.02

1. Find the probability that the next litter will produce at most six live pups.

2. compute the mean variance and the standard diviation of X interpet the mean in the contest of the problem.​

Answer 1

1. The probability of producing at most six live pups is the sum of the probabilities for X = 3, 4, 5, 6:

P(X ≤ 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.04 + 0.10 + 0.26 + 0.31 = 0.71

Therefore, the probability that the next litter will produce at most six live pups is 0.71.

2. The mean, variance, and standard deviation of the probability distribution are:

Mean (μ) = Σ(x * p(x)) = 3(0.04) + 4(0.10) + 5(0.26) + 6(0.31) + 7(0.22) + 8(0.05) + 9(0.02) = 5.68

Variance (σ^2) = Σ[(x - μ)^2 * p(x)] = (3 - 5.68)^2(0.04) + (4 - 5.68)^2(0.10) + (5 - 5.68)^2(0.26) + (6 - 5.68)^2(0.31) + (7 - 5.68)^2(0.22) + (8 - 5.68)^2(0.05) + (9 - 5.68)^2(0.02) = 1.7824

Standard Deviation (σ) = √σ^2 = √1.7824 = 1.335

The mean of the probability distribution is 5.68, which means that the expected number of live pups in a litter is 5.68. The variance of the probability distribution is 1.7824, which measures how spread out the distribution is. The standard deviation of the probability distribution is 1.335, which is the square root of the variance and also measures how spread out the distribution is.