Answer:
The minimum sample size needed for use of the normal approximation is 50.
Step-by-step explanation:
Suitability of the normal distribution:
In a binomial distribution with parameters n and p, the normal approximation is suitable is:
np >= 5
n(1-p) >= 5
In this question, we have that:
p = 0.9
Since p > 0.5, it means that np > n(1-p). So we have that:





The minimum sample size needed for use of the normal approximation is 50.
14 = M each on would be *t=5
U have to take a pic or the graph buddy
To see how much she saved, you multiply 24.90 x 10 ÷ 100 (or to simplify it 24.90 ÷ 10) you should get 2.49, which you subtract from the original and get $22.41
A) -18
B) 129
C) 40
D) -9
E) -8