We have the formula to compute the probability of having exactly k successed over n trials, given a probability p of success (and implicitly a probability 1-p of failure), which is
Now, the probability of at least 3 successes is the union of the following event: exactly three successes,exactly four successes and exactly five successes.
We can compute their probability and sum them:
So, the answer is about 36.79%
Answer:
CI ≈ (173.8 < μ < 196.2)
Step-by-step explanation:
We are told that laboratory tested twelve chicken eggs. Thus;
n = 12
Mean; x¯ = 185 mg
S.D; s = 17.6 mg
DF = n - 1 = 12 - 1 = 11
We have a 95% confidence level. Thus; α = 0.05
Since n < 30, we will use t-sample test.
Thus, from t-table attached at 95% Confidence level and DF = 11, we have;
t = 2.201
Thus,formula for Confidence interval is;
CI = (x¯ - t(s/√n)) < μ < (x¯ + t(s/√n))
CI = (185 - 2.201(17.6/√12)) < μ < (185 + 2.201(17.6/√12))
CI = (185 - 11.1825) < μ < (185 + 11.1825)
CI = (173.8175 < μ < 196.1825)
CI ≈ (173.8 < μ < 196.2)
