10. According to a survey, about 58% of families sit down for a fumily dinner at least four times per week. You ask 5 randomly chosen families whether they have a family dinner at least four times per week. a. Draw a histogram of the binomial distribution for the survey.
b. What is the most likely outcome of the survey?
c. What is the probability that at least 3 families have a family dinner four times per week?

Step-by-step explanation:

Use binomial probability:

P = nCr pʳ qⁿ⁻ʳ

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1−p).

Given n = 5, p = 0.58, and q = 0.42.

P(0) = ₅C₀ (0.58)⁰ (0.42)⁵⁻⁰ = 0.013

P(1) = ₅C₁ (0.58)¹ (0.42)⁵⁻¹ = 0.090

P(2) = ₅C₂ (0.58)² (0.42)⁵⁻² = 0.249

P(3) = ₅C₃ (0.58)³ (0.42)⁵⁻³ = 0.344

P(4) = ₅C₄ (0.58)⁴ (0.42)⁵⁻⁴ = 0.238

P(5) = ₅C₅ (0.58)⁵ (0.42)⁵⁻⁵ = 0.066

The most likely result is 3 families.

The probability of at least 3 families is:

0.344 + 0.238 + 0.066 = 0.648

because they could drop even lower..

a

step-by-step explanation:

u take $30 minus $24. the answer/original($30)