The probability that the sample mean is greater than 106.4 gallons is approximately 51.64%.
Given:
Mean (µ) = 107 gallons
Variance (σ²) = 49 gallons²
Sample size (n) = 46 racing cars
We need to find the probability that the sample mean (X) is greater than 106.4 gallons. In other words, we need to find P(X > 106.4).
According to the Central Limit Theorem, for a sufficiently large sample size, the distribution of the sample mean is approximately normal, with mean µ and variance σ²/n.
In this case, n = 46 is sufficiently large, so we can use the normal distribution to approximate the distribution of the sample mean.
The standard deviation (σ) is the square root of the variance, so σ = √σ² = √49 = 7 gallons.
To find P(X > 106.4), we can calculate the z-score of 106.4 and then use the z-table to find the probability that the z-score is greater than 106.4.
The z-score is calculated as:
z = (X - µ) / σ = (106.4 - 107) / 7 ≈ -0.086
Using the z-table, we find that the probability of a z-score greater than -0.086 is 0.5164.
Therefore, the probability that the sample mean would be greater than 106.4 gallons is approximately 0.5164 or 51.64%.