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The average fuel efficiency of U.S. light vehicles for 2005 was 21 mpg. If the standard deviation of the population was 2.9 and the gas ratings were normally distributed,

1) What is the probability that the mean mpg for a random sample of 25 light vehicles is under 20?

2) What is the probability that the mean mpg for a random sample of 25 light vehicles is between 20 and 25?

User Olga Botvinnik
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2 Answers

8 votes
8 votes

Here is a possible rewrite of the text in a more educational tone:

The mean mpg (miles per gallon) for a random sample of 25 light vehicles is a statistic that estimates the mean mpg for the population of all light vehicles. To find the probability of this statistic being under or between certain values, we need to use the concept of the sampling distribution of the mean. This is the distribution of all possible sample means from the population.

Since the gas ratings for the population are normally distributed, we can assume that the sampling distribution of the mean is also normally distributed. This means that we can use the normal distribution model to calculate probabilities for the sample mean.

To use the normal distribution model, we need two parameters: the mean and the standard deviation. The mean of the sampling distribution of the mean is equal to the population mean, which is given as 21. The standard deviation of the sampling distribution of the mean (also known as the standard error) is calculated by dividing the population standard deviation by the square root of the sample size. The population standard deviation is given as 2.9, and the sample size is 25.

Standard Error = Standard Deviation / √(Sample Size)

Standard Error = 2.9 / √25

Standard Error = 0.58

Now, we can use the normal distribution model to find probabilities for different values of the sample mean. To do this, we need to convert each value into a z-score, which measures how many standard errors away from the mean it is. The formula for z-score is:

z-score = (Sample Mean - Population Mean) / Standard Error

1) To find the probability that the sample mean is under 20, we need to find the z-score for 20 and then find the area to the left of that z-score on the normal curve.

z-score = (20 - 21) / 0.58

z-score ≈ -1.72

The area to the left of -1.72 on the normal curve represents the probability that the sample mean is under 20. We can use a standard normal distribution table or a calculator to find this area.

The probability that the sample mean is under 20 is approximately 0.0427, or 4.27%.

2) To find the probability that the sample mean is between 20 and 25, we need to find the z-scores for both 20 and 25 and then find the area between them on the normal curve.

z-score1 = (20 - 21) / 0.58 ≈ -1.72

z-score2 = (25 - 21) / 0.58 ≈ 6.90

The area between -1.72 and 6.90 on the normal curve represents the probability that the sample mean is between 20 and 25. We can find this area by subtracting the area to the left of -1.72 from the area to the left of 6.90.

Probability = Area to left of z-score2 - Area to left of z-score1

We can use a standard normal distribution table or a calculator to find these areas.

The probability that the sample mean is between 20 and 25 is approximately 1, as almost all of the area under the normal curve is between these two values.

Please note that these probabilities are approximations and may vary slightly depending on how you round your calculations.

User PMende
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10 votes
10 votes

Answer:

1) The probability that the mean mpg for a random sample of 25 light vehicles is 0.042341.

2) between 20 and 25 --> 21-25/2.9 = -1.38

Explanation:

Problem #1:

  • Using the z-score formula, z = (x-μ)/σ/n, where x is the raw score = 20 mpg,μ is the population mean = 21 mpg , σ is the population standard deviation = 2.9, n = random number of samples.

X < 20

  • = z = 20 - 21/2.9/√25
  • = z = -1/2.9/5
  • = z = -1.72414

Now

P-value from Z-Table:

P(x<20) = 0.042341

Problem #2:

21-25/2.9 = -1.38

User Eradicatore
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