Answer:
The required confidence interval is (3.068,4.732)
Step-by-step explanation:
Consider the provided information.
He plans to use a 90% confidence interval. He surveys a random sample of 50 students. The sample mean is 3.90 alcoholic drinks per week. The sample standard deviation is 3.51 drinks and wants to construct 90% confidence interval.
Thus, n=50,
=3.90 σ=3.51
Now find degree of freedom.

The confidence level is 90% and df=49
Therefore,


Now by using t distribution table look at 49 df and alpha level on 0.05.

Calculate SE as shown:


Now multiply 1.67653 with 0.4964
Therefore, the marginal error is: 1.67653 × 0.4964≈ 0.832
Now add and subtract this value in given mean to find the confidence interval.

Hence, the required confidence interval is (3.068,4.732)
Answer:
There is a 31/32 chance of having at least one boy
Answer:
No idea for this question
Let
T---------> <span>the tax rate on the parts in $
we know that
$18.75=$15+$3.50+$T
T=18.75-15-3.5
T=$0.25
so
$3.50----------> 100%
$0.25--------> x
x=0.25*100/3.50------> x=7.14%------> x=7.1%
the answer is
7.1%</span>