SOLUTION: A brisk walk at a 6 km per hour, burns an average of 300 calories per hour. If the standard deviation of the distribution is 8 calories, find the probability that a person who walk

Algebra ->  Test -> SOLUTION: A brisk walk at a 6 km per hour, burns an average of 300 calories per hour. If the standard deviation of the distribution is 8 calories, find the probability that a person who walk      Log On


   

Question 1193352: A brisk walk at a 6 km per hour, burns an average of 300 calories per hour. If the standard deviation of the distribution is 8 calories, find the probability that a person who walks an hour at the rate of 6 kilometers per hour will burn the following calories.
A.Less than 294 calories
B. Between 278 and 318 calories
Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
**1. Define Variables**
* Let X be the random variable representing the number of calories burned per hour.
* X follows a normal distribution with:
* Mean (μ) = 300 calories
* Standard deviation (σ) = 8 calories
**2. Standardize the Values**
* We'll use the z-score formula:
* z = (X - μ) / σ
**A. Less than 294 calories**
* Calculate the z-score for X = 294:
* z = (294 - 300) / 8 = -0.75
* Find the probability:
* P(X < 294) = P(Z < -0.75)
* Using a standard normal distribution table or a calculator, we find P(Z < -0.75) ≈ 0.2266
* **Therefore, the probability of burning less than 294 calories is approximately 0.2266 (or 22.66%).**
**B. Between 278 and 318 calories**
* Calculate the z-scores for X = 278 and X = 318:
* z1 = (278 - 300) / 8 = -2.75
* z2 = (318 - 300) / 8 = 2.25
* Find the probability:
* P(278 < X < 318) = P(-2.75 < Z < 2.25)
* P(-2.75 < Z < 2.25) = P(Z < 2.25) - P(Z < -2.75)
* Using a standard normal distribution table or a calculator:
* P(Z < 2.25) ≈ 0.9878
* P(Z < -2.75) ≈ 0.0030
* P(-2.75 < Z < 2.25) ≈ 0.9878 - 0.0030 = 0.9848
* **Therefore, the probability of burning between 278 and 318 calories is approximately 0.9848 (or 98.48%).**