I know this is probably too late but its 60%
Answer:
it is going to be slipper but the anwser is 2,200
Step-by-step explanation:
i did that by caucalting in my head
Answer:
You'll need $2,500 in sales this week to end up with $600 in income for the week.
Step-by-step explanation:
Let s represent the dollar amount of the total sales for the week.
The commission earned on such sales is 1/5, or 0.20, of s:
Commission: 0.20s
Then the correct expression for your weekly income is:
$100 + 0.20s = $600.
Solving for s, subtract $100 from both sides. Then 0.20s = $500, and through division we get s = $2,500.
You'll need $2,500 in sales this week to end up with $600 in income for the week.
We have:
n = 161
xbar = 32.8hg
s = 7.2hg
cl = 90% = 0.90
Since we have only the sample standard deviation s, we need to use the t-distribution.
The degrees of freedom DF = 161 - 1 = 160
α = (1 - 0.90)/2 = 0.05
If you look at these values in a t-distribution table you find t = 1.65
Now we can build the confidence interval:
xbar +/- (t · s /√n) = 32.8 +/- (1.65 · 7.2 / √161)
Therefore:
31.86 < μ < 33.74
In order to understand if these values are different from the ones you are given, let's calculate the confidence interval for the latest:
n = 12
xbar = 33.1hg
s = 2.7hg
31.7 < μ < 34.5
Calculate the error: (34.5 - 33.1) = (33.1 - 31.7) = 1.4
We know t · s /√n = 1.4
and we can solve for t:
t = 1.4 · √n / s = 1.4 · √12 / 2.7 = 1.7962
Looking at a t-distribution table, we find α = 0.05
which brings to a confidence level of 90%, which is the same for the previous part.
Since the sample size is big enough, we can use the normal distribution and the z-score:
Looking at a normal distribution table, we find z-score = 1.645, which is very similar to the t-value found previously. We don't know the population standard deviation, but for such a big sample the sample standard deviation is a good estimate, therefore:
<span>xbar +/- (z* · s /√n) = 32.8 +/- (1.645 · 7.2 / √161)
</span><span>31.8 < μ < 33.7</span>