Question

1. twenty - six percent of couples who plan to marry this year are planning destination weddings. in a random sample of 10 couples who plan to marry, find the probability that

a. exactly 6 couples will have a destination wedding

b. at least 6 couples will have a destination wedding

c. fewer than 5 couples will have a destination wedding

1. twenty - five percent of commercial airline accidents are caused by bad weather. if 300 commercial accidents are randomly selected, find the mean, variance, and standard deviation of the number of accidents caused by bad weather.

Explanation:

Problem 36 (Binomial Probability, \( n = 10 \), \( p = 0.26 \))

Part a: Exactly 6 couples

Step 1: Recall Binomial Formula

The binomial probability formula is \( P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \), where \( \binom{n}{k} = \frac{n!}{k!(n - k)!} \), \( n = 10 \), \( k = 6 \), \( p = 0.26 \), \( 1 - p = 0.74 \).

Step 2: Calculate Combination

\( \binom{10}{6} = \frac{10!}{6!4!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \).

Step 3: Substitute into Formula

\( P(X = 6) = 210 \times (0.26)^6 \times (0.74)^4 \).
Calculate \( (0.26)^6 \approx 0.0003089 \), \( (0.74)^4 \approx 0.3015 \).
Multiply: \( 210 \times 0.0003089 \times 0.3015 \approx 0.0195 \).

Part b: At least 6 couples (\( X \geq 6 \))

Step 1: Define "At Least 6"

\( P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \).

Step 2: Calculate Each Term

• \( P(X = 7) = \binom{10}{7} (0.26)^7 (0.74)^3 \)

\( \binom{10}{7} = 120 \), \( (0.26)^7 \approx 0.0000803 \), \( (0.74)^3 \approx 0.4052 \)
\( P(X = 7) \approx 120 \times 0.0000803 \times 0.4052 \approx 0.0039 \).

• \( P(X = 8) = \binom{10}{8} (0.26)^8 (0.74)^2 \)

\( \binom{10}{8} = 45 \), \( (0.26)^8 \approx 0.0000209 \), \( (0.74)^2 = 0.5476 \)
\( P(X = 8) \approx 45 \times 0.0000209 \times 0.5476 \approx 0.0005 \).

• \( P(X = 9) = \binom{10}{9} (0.26)^9 (0.74)^1 \)

\( \binom{10}{9} = 10 \), \( (0.26)^9 \approx 0.0000054 \), \( 0.74 \)
\( P(X = 9) \approx 10 \times 0.0000054 \times 0.74 \approx 0.00004 \).

• \( P(X = 10) = \binom{10}{10} (0.26)^{10} (0.74)^0 \)

\( \binom{10}{10} = 1 \), \( (0.26)^{10} \approx 0.0000014 \), \( (0.74)^0 = 1 \)
\( P(X = 10) \approx 1 \times 0.0000014 \times 1 \approx 0.0000014 \).

Step 3: Sum All Terms

\( P(X \geq 6) \approx 0.0195 + 0.0039 + 0.0005 + 0.00004 + 0.0000014 \approx 0.0239 \).

Part c: Fewer than 5 couples (\( X < 5 \))

Step 1: Define "Fewer Than 5"

\( P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \).

Step 2: Calculate Each Term

• \( P(X = 0) = \binom{10}{0} (0.26)^0 (0.74)^{10} \)

\( \binom{10}{0} = 1 \), \( (0.26)^0 = 1 \), \( (0.74)^{10} \approx 0.0498 \)
\( P(X = 0) \approx 1 \times 1 \times 0.0498 = 0.0498 \).

• \( P(X = 1) = \binom{10}{1} (0.26)^1 (0.74)^9 \)

\( \binom{10}{1} = 10 \), \( (0.74)^9 \approx 0.0673 \)
\( P(X = 1) \approx 10 \times 0.26 \times 0.0673 \approx 0.1750 \).

• \( P(X = 2) = \binom{10}{2} (0.26)^2 (0.74)^8 \)

\( \binom{10}{2} = 45 \), \( (0.26)^2 = 0.0676 \), \( (0.74)^8 \approx 0.0909 \)
\( P(X = 2) \approx 45 \times 0.0676 \times 0.0909 \approx 0.2759 \).

• \( P(X = 3) = \binom{10}{3} (0.26)^3 (0.74)^7 \)

\( \binom{10}{3} = 120 \), \( (0.26)^3 = 0.017576 \), \( (0.74)^7 \approx 0.1228 \)
\( P(X = 3) \approx 120 \times 0.017576 \times 0.1228 \approx 0.2552 \).

• \( P(X = 4) = \binom{10}{4} (0.26)^4 (0.74)^6 \)

\( \binom{10}{4} = 210 \), \( (0.26)^4 = 0.00456976 \), \( (0.74)^6 \approx 0.1660 \)
\( P(X = 4) \approx 210 \times 0.00456976 \times 0.1660 \approx 0.1593 \).

Step 3: Sum All Terms

\( P(X < 5) \approx 0.0498 + 0.1750 + 0.2759 + 0.2552 + 0.1593 \approx 0.9152 \).

Problem 37 (Binomial Mean, Variance, Standard Deviation)

Answer:

\( \approx 0.0195 \)

36b: