The purchases made by customers at a convenience store are normally distributed, with a
mean of $5.50 and a standard deviation of $3.50. What is the probability that a randomly chosen
customer makes a purchase between $2.00 and $9.00?
plug in your data at
http://davidmlane.com/hyperstat/z_table.html
To find the probability that a randomly chosen customer makes a purchase between $2.00 and $9.00, we can use the concept of the standard normal distribution.
1. First, we need to standardize the values ($2.00 and $9.00) using the formula for z-score:
z = (x - μ) / σ
where x is the value, μ is the mean, and σ is the standard deviation.
For $2.00:
z1 = ($2.00 - $5.50) / $3.50
For $9.00:
z2 = ($9.00 - $5.50) / $3.50
2. Calculate the z-scores:
z1 = -1.00
z2 = 1.00
3. Look up the corresponding probabilities for the standardized z-scores from the standard normal distribution table.
The probability for a z-score of -1.00 is 0.1587.
The probability for a z-score of 1.00 is 0.8413.
4. Subtract the probability corresponding to the lower z-score from the probability corresponding to the higher z-score:
P(2.00 < x < 9.00) = P(z1 < z < z2) = P( -1.00 < z < 1.00) = P(z < 1.00) - P(z < -1.00)
P(2.00 < x < 9.00) = 0.8413 - 0.1587
5. Calculate the final probability:
P(2.00 < x < 9.00) ≈ 0.6826
Therefore, the probability that a randomly chosen customer makes a purchase between $2.00 and $9.00 is approximately 0.6826, or 68.26%.