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Chapter 6: Problem 56

The average birth weight of elephants is 230 pounds. Assume that the distribution of birth weights is Normal with a standard deviation of 50 pounds. Find the birth weight of elephants at the 95 th percentile.

Short Answer

Expert verified
The birth weight of elephants at the 95th percentile is approximately 312.25 pounds.

Step by step solution

01

- Define the knowns

In this exercise, the known variables are the mean \(\mu = 230\) pounds and the standard deviation \(\sigma=50\) pounds. The desired percentile to find is 95% or \(0.95 \) in decimal form.
02

- Standardize the variable

Next, we define the z-score which corresponds to the 95th percentile. Z-score is the number of standard deviations from the mean that a certain data point is. We find this usually from a Z-table or a statistical calculator. In this case, the value is approximately \(1.645 \).
03

- Apply the Z-score formula

The Z-score formula is \(z = (X - \mu) / \sigma \), where \(X\) is the data point. We will rearrange this formula to solve for \(X\), the unknown variable representing the 95th percentile birth weight. Thus, the formula becomes \(X = z *\sigma + \mu\).
04

- Calculate birth weight

Substitute the known values into the rearranged Z-score formula. \(Z = 1.645\), \(\sigma = 50\) and \(\mu = 230\). So \(X = 1.645*50 + 230 \).

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Most popular questions from this chapter

Assume a standard Normal distribution. Draw a separate, well-labeled Normal curve for each part. a. Find the \(z\) -score that gives a left area of \(0.9774\). b. Find the \(z\) -score that gives a left area of \(0.8225\).

According to the College Board, the mean quantitative SAT score for male college-bound high school seniors in one year was \(530 .\) SAT scores are approximately Normally distributed with a population standard deviation of \(100 .\) What is the SAT score at the 96 th percentile for male college-bound seniors?

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