Answer: 1) 0.10
2) 0.60
3) 0.20
4) 0.10
<u>Step-by-step explanation:</u>
The total frequency is 20+120+40+20 = 200. This means they ran the experiment 200 times. The probability distribution is calculated by the satisfactory number of outcomes (frequency) divided by the total number of experiments/outcomes (total frequency):
![\begin{array}{c|c||lc}\underline{x}&\underline{f}&\underline{f\div 200}&\underline{\text{Probability Distribution}}\\1&20&20\div200=&0.10\\2&120&120\div 200=&0.60\\3&40&40\div 200=&0.20\\4&20&20\div 200=&0.10\end{array}\right]](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bc%7Cc%7C%7Clc%7D%5Cunderline%7Bx%7D%26%5Cunderline%7Bf%7D%26%5Cunderline%7Bf%5Cdiv%20200%7D%26%5Cunderline%7B%5Ctext%7BProbability%20Distribution%7D%7D%5C%5C1%2620%2620%5Cdiv200%3D%260.10%5C%5C2%26120%26120%5Cdiv%20200%3D%260.60%5C%5C3%2640%2640%5Cdiv%20200%3D%260.20%5C%5C4%2620%2620%5Cdiv%20200%3D%260.10%5Cend%7Barray%7D%5Cright%5D)
Answer:
C 16
Step-by-step explanation:
Using ratios
3:12
12/3=4
4:x
4*4=16
x=16
4:16
hope this helps:)
So with the 10% discount, thats 45$, adding tax, that's <span>$47.70.</span>
The valid conclusions for the manager based on the considered test is given by: Option
<h3>When do we perform one sample z-test?</h3>
One sample z-test is performed if the sample size is large enough (n > 30) and we want to know if the sample comes from the specific population.
For this case, we're specified that:
- Population mean =
= $150 - Population standard deviation =
= $30.20 - Sample mean =
= $160 - Sample size = n = 40 > 30
- Level of significance =
= 2.5% = 0.025 - We want to determine if the average customer spends more in his store than the national average.
Forming hypotheses:
- Null Hypothesis: Nullifies what we're trying to determine. Assumes that the average customer doesn't spend more in the store than the national average. Symbolically, we get:

- Alternate hypothesis: Assumes that customer spends more in his store than the national average. Symbolically

where
is the hypothesized population mean of the money his customer spends in his store.
The z-test statistic we get is:

The test is single tailed, (right tailed).
The critical value of z at level of significance 0.025 is 1.96
Since we've got 2.904 > 1.96, so we reject the null hypothesis.
(as for right tailed test, we reject null hypothesis if the test statistic is > critical value).
Thus, we accept the alternate hypothesis that customer spends more in his store than the national average.
Learn more about one-sample z-test here:
brainly.com/question/21477856