Answer: 0.1457
Step-by-step explanation:
Let p be the population proportion.
Given: The proportion of Americans who are afraid to fly is 0.10.
i.e. p= 0.10
Sample size : n= 1100
Sample proportion of Americans who are afraid to fly =
We assume that the population is normally distributed
Now, the probability that the sample proportion is more than 0.11:
![P(\hat{p}>0.11)=P(\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}>\dfrac{0.11-0.10}{\sqrt{\dfrac{0.10(0.90)}{1100}}})\\\\=P(z>\dfrac{0.01}{0.0090453})\ \ \ [\because z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}} ]\\\\=P(z>1.1055)\\\\=1-P(z\leq1.055)\\\\=1-0.8543=0.1457\ \ \ [\text{using z-table}]](https://tex.z-dn.net/?f=P%28%5Chat%7Bp%7D%3E0.11%29%3DP%28%5Cdfrac%7B%5Chat%7Bp%7D-p%7D%7B%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%3E%5Cdfrac%7B0.11-0.10%7D%7B%5Csqrt%7B%5Cdfrac%7B0.10%280.90%29%7D%7B1100%7D%7D%7D%29%5C%5C%5C%5C%3DP%28z%3E%5Cdfrac%7B0.01%7D%7B0.0090453%7D%29%5C%20%5C%20%5C%20%5B%5Cbecause%20z%3D%5Cdfrac%7B%5Chat%7Bp%7D-p%7D%7B%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%20%5D%5C%5C%5C%5C%3DP%28z%3E1.1055%29%5C%5C%5C%5C%3D1-P%28z%5Cleq1.055%29%5C%5C%5C%5C%3D1-0.8543%3D0.1457%5C%20%5C%20%5C%20%5B%5Ctext%7Busing%20z-table%7D%5D)
Hence, the probability that the sample proportion is more than 0.11 = 0.1457
The answer is 116631.6 if rounded it is 116632
We need to figure out how much string would be left, after taking away the first two pieces.
We know that the first piece is 20 inches long, so we can say that there is 52-20 inches left, or 32 inches.
The second piece is between 12 and 18 inches, meaning that there would be between 32-12 and 32-18 inches left for the third piece, or 20 and 14 inches. This means that the third piece would be at least 14 inches long, but no more than 20, since we don’t have more string than that (20+12+20=52, and 20+14+18=52)
So we can say that x is greater or equal to 14, but less than or equal to 20, or:
14<=x<=20 (“<=“ is written like a normal “<“ sign with a line _ right under it)
X ≤ 1 and x ≥ 3
Hope this helps :)