Answer:
The standard deviation is 4.83 inches.
Explanation:
We are given that the average yearly snowfall in Chillyville is Normally distributed with a mean of 55 inches.
The snowfall in Chillyville exceeds 60 inches in 15% of the years, we have to find the standard deviation.
Let X = <u><em>the average yearly snowfall in Chillyville</em></u>.
The z-score probability distribution for the normal distribution is given by;
Z = 
~ N(0,1)
where, 
= mean amount of rainfall = 55 inches
= standard deviation
Now, it is stated that the snowfall in Chillyville exceeds 60 inches in 15% of the years, that means;
P(X > 60 inches) = 0.15
P( > 
) = 0.15
P(Z > ) = 0.15
In the z table, the critical value of z that represents the top 15% of the area is given as 1.0364, that means;
= 4.83 inches
Hence, the standard deviation is 4.83 inches.
Answer:
- <u><em>D. No, because $100,000 is much greater than the values used in the experiment.</em></u>
<u><em></em></u>
Explanation:
<em>Correlations</em>, when have strong correlation coefficients, which r = 0.9 is, may be good predictors within the limits of the range of the data.
Trying to extrapolate the <em>linear relationship</em> between the variables, <em>x = advertising spending and y= product sales</em>, way beyond the limits of the data used for the study, is too risky, because the data may be linear just for some stages (ranges) but behave very different in other ranges.
As, the option D. states, <em>$100,000 is much greater that the values used in the experiment</em>; hence, the correlation would likely would not be a good predictor for that input.
Hello. You did not add the graphics to which this question refers, which makes it impossible for your question to be answered. However, I created a model chart that represents the situation shown in the question and that can help you. This chart is attached in the figure below.
Notice, in the graph that the more students attended more character education classes, the greater their acts of foolishness. On the other hand, these acts tend to decrease as students have attended less classes, reaching a very low result for students who have not attended any class of this type of education.
I think the third option seem correct just off the top of my head. Not sure though
