Answer:
The probability of getting a sample with 80% satisfied customers or less is 0.0125.
Step-by-step explanation:
We are given that the results of 1000 simulations, each simulating a sample of 80 customers, assuming there are 90 percent satisfied customers.
Let
= <u><em>sample proportion of satisfied customers</em></u>
The z-score probability distribution for the sample proportion is given by;
Z =
~ N(0,1)
where, p = population proportion of satisfied customers = 90%
n = sample of customers = 80
Now, the probability of getting a sample with 80% satisfied customers or less is given by = P(
80%)
P(
80%) = P(
) = P(Z
-2.24) = 1 - P(Z < 2.24)
= 1 - 0.9875 = <u>0.0125</u>
The above probability is calculated by looking at the value of x = 2.24 in the z table which has an area of 0.9875.
<span>The answer is: the least amount is 105.35 and the greatest amount is 105.44. If the number after the one you want to round is 5 or bigger, you need to round up the number. For instance, 105.35 is rounded to 105.4 which is equal to 105.40. If the number after the one you want to round is smaller than 5, you need to round down the number. For instance, 105.44 is rounded to 105.4 which is equal to 105.40.</span>
There are a total of 30 integers and 10 of these are divisible by 3 (3, 6, 9, 12, 15, 18, 21, 24, 27, and 30). So the probability of getting an integer divisible by 3 is

= 1/3. The answer is letter C.