Using the z-distribution, as we have the standard deviation for the population, it is found that the smallest sample size required to obtain the desired margin of error is of 77.
<h3>What is a z-distribution confidence interval?</h3>
The confidence interval is:

In which:
is the sample mean.
is the standard deviation for the population.
The margin of error is given by:

In this problem, we have that the parameters are given as follows:
.
Hence, solving for n, we find the sample size.






Rounding up, the smallest sample size required to obtain the desired margin of error is of 77.
More can be learned about the z-distribution at brainly.com/question/25890103
Answer:
The probability is;
0.0000003645
Step-by-step explanation:
The probability of packages being early p is 90% = 0.9
The probability of packages being late q will be 1-p = 1-0.9 = 0.1
So the probability of 2 out of 10 random late will be subject to Bernoulli approximation of the Binomial theorem
That will be;
P(X = 2) = 10 C 2 0.9^2 0.1^8
= 0.0000003645
2 liters of paint / 1/3 liter per wall = 2 x 3/1 = 6/1 = 6 walls
1980 should be ur answer BUT if you want it estimated it would be 1800
Answer: 345.977 m^3
Step-by-step explanation: