A local bank surveyed the status of ASU student accounts and found that the average overdraft was $21.22 with a standard deviation of $5.49. If the distribution is normal, find the probability of a student being overdrawn by more than $18.75.

the probability of a student being overdrawn by more than $18.75 is 0.674

Step-by-step explanation:

Since the bank overdrafts of ASU student accounts are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = bank overdraft of Asu students.

µ = mean

σ = standard deviation

From the information given,

µ = $21.22

σ = $5.49

We want to find the probability of a student being overdrawn by more than $18.75. It is expressed as

P(x > 18.75) = 1 - P(x ≤ 18.75)

For x = 18.75,

z = (18.75 - 21.22)/5.49 = - 0.45

Looking at the normal distribution table, the probability corresponding to the z score is 0.326

Therefore,

P(x > 18.75) = 1 - 0.326 = 0.674

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

step-by-step explanation: