The mean amount of money withdrawn from a bank ATM on a particular day was $120. If Lauren withdrew $80 and her z-score was -3.2, what was the standard deviation for the amount withdrawn that day?
z-score = (given data - mean)/ sd
-3.2 = (80-120)/sd
-3.2sd = -40
sd = 12.5
To find the standard deviation, we need to use the formula for z-score:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the value
- μ is the mean
- σ is the standard deviation
In this case, we know that Lauren's z-score is -3.2 and her withdrawal amount x is $80. The mean withdrawal amount μ is given as $120.
So, let's rearrange the formula to solve for the standard deviation σ:
σ = (x - μ) / z
Substituting the given values:
σ = ($80 - $120) / -3.2
Simplifying:
σ = -$40 / -3.2
σ ≈ $12.50
Therefore, the standard deviation for the amount withdrawn that day is approximately $12.50.