The following were the recorded birth weights for babies born July 16, 2011: 8.1 lbs., 6.0 lbs., 4.7 lbs., 6.9 lbs., 5.6 lbs., 7
1 answer:
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Using the normal distribution, it is found that:
- 3 - a) The 40th percentile of the height of Dinaric Alps distribution for men is of 72.2 inches.
- 3 - b) The minimum height of man in the Dinaric Alps that would place him in the top 10% of all heights is of 76.84 inches.
- 4 - a) The 25th percentile for the math scores was of 71.6 inches.
- 4 - b) The 75th percentile for the math scores was of 78.4 inches.
<h3>Normal Probability Distribution </h3>
In a <em>normal distribution </em>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.
Question 3:
- The mean is of 73 inches, hence
.
- The standard deviation is of 3 inches, hence
.
Item a:
The 40th percentile is X when Z has a p-value of 0.4, so <u>X when Z = -0.253</u>.
The 40th percentile of the height of Dinaric Alps distribution for men is of 72.2 inches.
Item b:
The minimum height is the 100 - 10 = 90th percentile is X when Z has a p-value of 0.9, so <u>X when Z = 1.28</u>.
The minimum height of man in the Dinaric Alps that would place him in the top 10% of all heights is of 76.84 inches.
Question 4:
- The mean score is of 75, hence
.
- The standard deviation is of 5, hence
.
Item a:
The 25th percentile is X when Z has a p-value of 0.25, so <u>X when Z = -0.675</u>.
The 25th percentile for the math scores was of 71.6 inches.
Item b:
The 75th percentile is X when Z has a p-value of 0.25, so <u>X when Z = 0.675</u>.
The 75th percentile for the math scores was of 78.4 inches.
To learn more about the normal distribution, you can take a look at brainly.com/question/24663213