Answer:
Probability that ten or more homeowners would state a preference for either brand A or brand B = 0.302
Step-by-step explanation:
Given - A manufacturer of floor wax has developed two new brands, A and B, which she wishes to subject to homeowners’ evaluation to determine which of the two is superior. Both waxes, A and B, are applied to floor surfaces in each of 15 homes. Assume that there is actually no difference in the quality of the brands.
To find - What is the probability that ten or more homeowners would state a preference for either brand A or brand B?
Proof -
Given that,
There is actually no difference in the quality of the brands.
⇒P(A) = = 0.5
P(B) = = 0.5
Now,
We know that,
Binomial distribution is equals to
B(n, x) = ⁿCₓ (p)ˣ (1 - p)ⁿ⁻ˣ
Now,
P(X(A) ≥ 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)
= ¹⁵C₁₀ (0.5)¹⁰ (0.5)⁵ + ¹⁵C₁₁ (0.5)¹¹ (0.5)⁴ + ¹⁵C₁₂ (0.5)¹² (0.5)³ + ¹⁵C₁₃ (0.5)¹³ (0.5)² + ¹⁵C₁₄ (0.5)¹⁴ (0.5)¹ + ¹⁵C₁₅ (0.5)¹⁵ (0.5)⁰
= 0.092 + 0.042 + 0.014 + 0.003 + 0.0005 + 0.00005
= 0.151
⇒P(X(A) ≥ 10) = 0.151
As given Quality of both brands are same ,
So, the probability of both the brands are equal
⇒P(X(B) ≥ 10) = 0.151
Now,
Probability that ten or more homeowners would state a preference for either brand A or brand B = P(X(A) ≥ 10) + P(X(A) ≥ 10)
= 0.151 + 0.151
= 0.302
∴ we get
Probability that ten or more homeowners would state a preference for either brand A or brand B = 0.302
