The amount that airlines speed on food per passenger is normally distributed with means $8.00 and standard deviation $2.00. What is the probability that a randomly selected airline spends more than $12.50?
To find the probability that a randomly selected airline spends more than $12.50, we need to use the standard normal distribution.
Step 1: Standardize the given value.
The formula to standardize a value is:
z = (x - μ) / σ
Where:
x = given value
μ = mean
σ = standard deviation
In this case, the value is $12.50, the mean is $8.00, and the standard deviation is $2.00.
Plugging in these values into the formula, we have:
z = (12.50 - 8.00) / 2.00
Calculating this, we get:
z = 4.50 / 2.00
z = 2.25
Step 2: Find the probability.
Since we want to find the probability that a randomly selected airline spends more than $12.50, we need to find the area under the standard normal curve corresponding to the z-value of 2.25. This area represents the probability.
To find this probability, we can use a table of the standard normal distribution or use a calculator with a normal distribution function. Let's use a standard normal distribution table for this explanation.
Looking up the z-value of 2.25 in the standard normal distribution table, we find that the corresponding probability is approximately 0.9878.
Step 3: Interpret the result.
Therefore, the probability that a randomly selected airline spends more than $12.50 is approximately 0.9878 or 98.78%.
In summary, the probability that a randomly selected airline spends more than $12.50 is approximately 0.9878 or 98.78%.