Respuesta :
Answer:
a) Mean of leaked tanks in a sample size of 15 = 3.75
b) Probability that 10 or more of the tanks out of 15 will leak = 0.000795
c) 0.034
Step-by-step explanation:
a) Mean number of leaking tanks in a sample size = np
n = sample size
p = probability of a leagage
Mean = 0.25× 15 = 3.75
b) The probability that 10 or more of the tanks leak
This is a binomial distribution problem
Binomial distribution function is represented by
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
n = total number of sample spaces = 15 tanks
x = Number of successes required = number of leaked tanks expected
p = probability of success = probability of a tank leaking = 0.25
q = probability of failure = 1 - p = 0.75
P(X ≥ 10) = P(X=10) + P(X=11) + P(X=12) + P(X=13) + P(X=14) + P(X=15)
P(X ≥ 10) = 0.000795
c) For a sample size of 1000, we first investigate if it follows a normal distribution as it satisfies the two required conditions for a normal distribution
np = 0.25 × 1000 = 250 > 10
np(1-p) = 1000 × 0.25 × 0.75 = 187.5 > 10
Mean = np = 250
Standard deviation = √[np(1-p)] = √(187.5) = 13.70
The probability that at least 275 of these tanks are leaking = P(x ≥ 275)
We first standardize 275
The standardized score for a value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (275 - 250)/13.7 = 1.825
P(x ≥ 275) = P(z ≥ 1.825)
We'll use data from the normal probability table for these probabilities
P(x ≥ 275) = P(z ≥ 1.825) = P(z ≤ - 1.825) = 0.034
