The funds dispensed at the ATM machine located near the checkout line at the Kroger's in Union Kentucky, follows a normal probability distribution with mean $4200 per day and a standard deviation of $720 a day. The machine is programmed to notify the nearby bank that if the amount dispensed is very low (less than $2500) or very high ($6,000).
a. What percent of the days will the bank be notified because the amount dispensed is very low?
b. What percent of the time will the bak be notified because the amount is very high?
c. What percent of the time will tbe bank not be notified regarding the amount of funds being dispensed?

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fichoh fichoh · Answer 1 · 2020-08-08T04:40:46+02:00

Answer:

0.00914 ; 0.0062 ; 0.9847

Explanation:

Given the following :

Normal distribution :

Mean(m) = $4200

Standard deviation (sd) = 720

Amount VERY low = < 2500

Amount very high = > 6000

a. What percent of the days will the bank be notified because the amount dispensed is very low?

x < 2500

Finding the z-score :

Z = (x - mean) / standard deviation

Z = (2500 - 4200) / 720

Z = - 1700 / 720 = −2.361111

P(z < −2.361111)

Locating −2.361111 on the z- distribution table

-2.3 under 0.06 = 0.00914

P(z < −2.361111) = 0.00914

B) What percent of the time will the bank be notified because the amount is very high?

x > 6000

Finding the z-score :

Z = (x - mean) / standard deviation

Z = (6000 - 4200) / 720

Z = 1800 / 720 = 2.5

P(z > 2.5)

Locating 2.5 on the z- distribution table = 0.9938

P(z > 2.5) = 1 - 0.9938 = 0.0062

c. What percent of the time will the bank not be notified regarding the amount of funds being dispensed?

P(2500 < X < 6000)

P(-2.36 < z < 0) + P(0 < z < 2.5)

= 0.4909 + 0.4938

= 0.9847