Normal Distribution Argumentative: Challenge and Solution Essay Example
Normal Distribution Argumentative: Challenge and Solution Essay Example

Normal Distribution Argumentative: Challenge and Solution Essay Example

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  • Pages: 2 (314 words)
  • Published: June 26, 2018
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The objective is to determine the likelihood of selecting a measurement between 19 and 31 from a normally distributed population with an average of 25 and a variance of 9. In order to accomplish this, we must convert the values 19 and 31 into their respective z values using the transformation formula for X to z (where µ = 25 and ?^2 = 9). The calculations demonstrate that z19 = (19 - 25) / v9 = -2 and z31 = (31 - 25) / v9 = +2. By determining the area between z = ±2, we can establish the desired probability.

4772) = 0.9554. Therefore, the chance of a randomly chosen measurement falling between 19 and 31 is approximately 0.95. This probability applies to both the X values and the z values. The area between 19 and 31, as well as the area between -2 and +2

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, is represented by a standard normal curve. Admission into a specific university relies on a nationwide exam with scores that follow a normal distribution having a mean of 500 and a standard deviation of 100. Tom's goal is to get accepted into this university, and he understands that he needs to outperform at least 70% of the test-takers.

Tom received a score of 585 on the test, and the question is whether this score meets the university's admission criteria. To evaluate this, we can utilize the random variable x to represent scores. The distribution of x follows a normal distribution with an average of 500 and a standard deviation of 100. The area beneath the normal curve corresponds to the number of students who took the test.

By multiplying the areas

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under the curve by 100, we can express them as percentages. For instance, if x equals 585, we can convert it to z with the formula z = (x - 500) / 100. This conversion results in z = 0.85. The proportion P of students who scored below 585 can be determined by calculating the area to the left of z = 0.85. This calculation yields P = 0.8023 or equivalently, P = 80%.

23% of the students who took the test will be admitted to this University, and Tom scored better than 80.

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