-1 <span>± </span><span><span>√61 / 6. Negative 1, plus or minus square root of 61 divide by 6</span></span>
To solve this problem, we make use of the z statistic.
What we have to do here is to find for the z score using the given data and
from the standard probability tables for z, we locate the proportion of the
fish longer than 12 inches.
The formula for calculating the z score is:
z = (x – μ) / σ
where,
x = the sample value = 12 inches
μ = the sample mean = 11 inches
σ = 2 inches
so,
z = (12 – 11) / 2
z = 1 / 2
z = 0.5
Since we are looking for the values greater than 12, so
this is a right tailed test. Using the tables, the value of p at this value of
z is:
p (z = 0.5) = 0.3085
Therefore there is a 30.85% probability that the fish is
longer than 12 inches.
Answer:
We are 95% confident that the true proportion of TV audience is between 65.15% and 65.85%
Step-by-step explanation:
-From the given information, 
.
-We calculate the confidence interval using this value at 95% confidence level:
![CI=\hat p\pm z \sqrt{\frac{\hat p(1-\hat p)}{n}}\\\\\\=0.65\pm 1.96\times \sqrt{\frac{0.65\times 0.35}{12000}}\\\\\\=0.65\pm 0.0085\\\\\\=[0.6415,0.6585]](https://tex.z-dn.net/?f=CI%3D%5Chat%20p%5Cpm%20z%20%5Csqrt%7B%5Cfrac%7B%5Chat%20p%281-%5Chat%20p%29%7D%7Bn%7D%7D%5C%5C%5C%5C%5C%5C%3D0.65%5Cpm%201.96%5Ctimes%20%5Csqrt%7B%5Cfrac%7B0.65%5Ctimes%200.35%7D%7B12000%7D%7D%5C%5C%5C%5C%5C%5C%3D0.65%5Cpm%200.0085%5C%5C%5C%5C%5C%5C%3D%5B0.6415%2C0.6585%5D)
So, the 95% confidence interval is (0.6515,0.6585).
Hence, we are 95% confident that the true proportion of TV audience is between 65.15% and 65.85%.
Answer:
yes
Step-by-step explanation:
