Question 1209960
Let's solve this problem step-by-step.

**Understanding the Problem**

This is a binomial probability problem because:

* There are a fixed number of trials (n = 5).
* Each trial has only two possible outcomes (nonfatal or fatal).
* The probability of success (nonfatal) is constant (p = 0.55).
* The trials are independent.

**Given Information**

* Probability of nonfatal shark attack (success), p = 0.55
* Probability of fatal shark attack (failure), q = 1 - p = 1 - 0.55 = 0.45
* Number of trials (sample size), n = 5

**Formula for Binomial Probability**

The probability of getting exactly k successes in n trials is:

P(X = k) = (nCk) * p^k * q^(n-k)

where:

* nCk = n! / (k! * (n-k)!) (the number of combinations of n items taken k at a time)

**a) Probability All Five Shark Attacks Are Nonfatal (k = 5)**

P(X = 5) = (5C5) * (0.55)^5 * (0.45)^(5-5)
P(X = 5) = 1 * (0.55)^5 * (0.45)^0
P(X = 5) = (0.55)^5 * 1
P(X = 5) ≈ 0.0503284375

**b) Probability Three or More of the Five Shark Attacks Are Nonfatal (k ≥ 3)**

We need to calculate P(X = 3), P(X = 4), and P(X = 5), then add them up.

* P(X = 3) = (5C3) * (0.55)^3 * (0.45)^(5-3) = 10 * (0.55)^3 * (0.45)^2 ≈ 0.275653125
* P(X = 4) = (5C4) * (0.55)^4 * (0.45)^(5-4) = 5 * (0.55)^4 * (0.45)^1 ≈ 0.2058890625
* P(X = 5) ≈ 0.0503284375 (calculated in part a)

P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5)
P(X ≥ 3) ≈ 0.275653125 + 0.2058890625 + 0.0503284375
P(X ≥ 3) ≈ 0.531870625

**c) Expected Number of Nonfatal Shark Attacks (Mean)**

For a binomial distribution, the expected number (mean) is:

* μ = n * p
* μ = 5 * 0.55
* μ = 2.75

**d) Standard Deviation of the r-Probability Distribution**

For a binomial distribution, the standard deviation is:

* σ = √(n * p * q)
* σ = √(5 * 0.55 * 0.45)
* σ = √(1.2375)
* σ ≈ 1.11242528

**Answers:**

a)  Approximately 0.0503
b)  Approximately 0.5319
c)  2.75
d)  Approximately 1.1124