The true highway mileage, Y, of a specific brand of hybrid vehicle has a normal distribution with a mean of 42 and a standard deviation of 1.5. Find the probability that a randomly selected hybrid vehicle gets more than 38 miles to the gallon.
μ=42
σ=1.5
so
Z(38)=(38-μ)/σ
Look up a normal distribution table for the probability of exceedingZ(38), which is the same as the probability of not exceeding -Z(38).
Post your answer for checking if you wish.
.9961
.0039 or 39%
To find the probability that a randomly selected hybrid vehicle gets more than 38 miles to the gallon, we need to calculate the area under the normal distribution curve to the right of 38.
To do this, we can use the standard normal distribution, which has a mean of 0 and a standard deviation of 1.
First, we need to standardize the value of 38 using the formula: z = (x - μ) / σ
where z is the z-score, x is the observed value, μ is the mean, and σ is the standard deviation.
In this case, x = 38, μ = 42, and σ = 1.5. Plugging in the values, we get:
z = (38 - 42) / 1.5
z = -4 / 1.5
z = -2.67
Now, we can find the probability using the standard normal distribution table or a calculator. The standard normal distribution table provides the cumulative probability up to a given z-score.
Using the table or a calculator, we find that the cumulative probability associated with a z-score of -2.67 is approximately 0.0039.
Since we are interested in the probability of the vehicle getting more than 38 miles to the gallon, we need to find the area in the right tail of the curve. This is equal to 1 minus the cumulative probability.
Therefore, the probability that a randomly selected hybrid vehicle gets more than 38 miles to the gallon is approximately 1 - 0.0039 = 0.9961 or 99.61%.