Using the normal distribution, it is found that 3.59% of sixth-graders earn a score of at least 93 on the reading test.
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Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula, which is given by:
- is the mean.
- is the standard deviation.
- It measures how many standard deviations the measure is from the mean. Each z-score has a p-value associated with it.
- This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
- Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
- On the reading test, the mean is of 75, thus .
- The standard deviation is of 10, thus .
- The proportion who scored above 93 is <u>1 subtracted by the p-value of Z when X = 93</u>, thus:
has a p-value of 0.9641.
1 - 0.9641 = 0.0359.
0.0359 x 100% = 3.59%
3.59% of sixth-graders earn a score of at least 93 on the reading test.
A similar problem is given at brainly.com/question/24663213
Solution: g + 1 ≠ 0,
Geometric figure: Hyperbola (on graph)
Answer:
8 hours
$114 - $50 and get $64 so $8 divided by $64 you’ll get 8