At the end of the Halloween Festival the organizers estimated that a family of participants spent an average $45.00 with a standard deviation of $12.00.
What’s the probability that the mean amount spent will be between $20 and $30?
To calculate the probability that the mean amount spent will be between $20 and $30, we need to standardize the values using the z-score formula and then use a standard normal distribution table or a calculator to find the probability.
The z-score formula is given by:
z = (x - μ) / (σ / sqrt(n))
Where:
- x is the value we want to standardize ($20 or $30 in this case),
- μ is the population mean ($45.00),
- σ is the population standard deviation ($12.00),
- sqrt(n) is the square root of the sample size (assumed to be equal to 1 in this case since we don't have the sample size, but it cancels out in the calculation).
First, let's calculate the z-scores for $20 and $30:
For $20:
z1 = (20 - 45) / (12 / sqrt(1)) = -25 / 12 = -2.08
For $30:
z2 = (30 - 45) / (12 / sqrt(1)) = -15 / 12 = -1.25
Next, we need to find the cumulative probabilities associated with these z-scores.
Using a standard normal distribution table or a calculator, we find:
For z1 = -2.08, the cumulative probability is approximately 0.0192 (or 1.92%).
For z2 = -1.25, the cumulative probability is approximately 0.1056 (or 10.56%).
To find the probability that the mean amount spent will be between $20 and $30, we subtract the cumulative probability associated with z2 from the cumulative probability associated with z1:
Probability = P(z1 <= z <= z2) = P(z <= z2) - P(z <= z1)
= 0.1056 - 0.0192
= 0.0864
≈ 8.64%
Therefore, the probability that the mean amount spent will be between $20 and $30 is approximately 8.64%.