I have four bills that need to be paid this week. The electric bill is 182.17 the water bill is 92.55 the cell phone bill is 225
2 answers:
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Using the normal distribution, it is found that:
1. His z-score was of Z = -1.88.
2. There is a 0.0301 = 3.01% probability that a randomly selected person has a smaller E/A ratio than the patient in question 1.
3. Z-score of z = 1.85, there is a 0.0322 = 3.22% probability that a randomly selected patient has a higher E/A ratio.
4. Due to the higher absolute value of the z-score, the first patient had a more extraordinary result.
<h3>Normal Probability Distribution</h3>In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem:
- The mean is of
.
- The standard deviation is of
.
Item 1:
Considering his ratio, we have that X = 0.73, hence:
His z-score was of Z = -1.88.
Item 2:
The probability is the <u>p-value of Z = -1.88</u>, hence, there is a 0.0301 = 3.01% probability that a randomly selected person has a smaller E/A ratio than the patient in question 1.
Item 3:
Score of X = 1.96, hence:
The probability is 1 subtracted by the p-value of Z = 1.85, hence, 1 - 0.9678 = 0.0322 = 3.22% probability that a randomly selected patient has a higher E/A ratio.
Item 4:
Due to the higher absolute value of the z-score, the first patient had a more extraordinary result.
More can be learned about the normal distribution at brainly.com/question/24663213
Answer:
We are given the correlation between height and weight for adults is 0.40.
We need to find the proportion of the variability in weight that can be explained by the relationship with height.
We know that coefficient of determination or R-square measures the proportion or percent of variability in dependent variable that can be explained by the relationship with independent variable. There the coefficient of determination is given below:
Therefore, the 0.16 or 16% of the variability in weight can be explained by the relationship with height