Help Save & Exit Submit 15 probability distribution with a mean of $4,200 per da

Question

Help Save & Exit Submit 15 probability distribution with a mean of $4,200 per day and a standard deviation of $720 per day. The machine is programmed to notify the nearby bank if the amount dispensed is very low (less than $2.500) or very high (more than $6,000) ated line at the Krogers in Union, Kentucky, foilows a normal 338 oins wiesia plere ene tnerbi r brs a. What percent of the days will the bank be notified because the amount dispensed is very low? (Round zscore computation to 2 decimal places and your final answer to 2 decimal places.) ey o References b. What percent of the time will the bank be notified because the amount dispensed is high? (Round -score computation to 2 decimal places and your final answer to 2 decimal places.) c. What percent of the time will the bank not be notified regarding the amount of funds dispersed? (Round your answer to 2 decimal

Explanation / Answer

Mean = $4200

Standard deviation = $720

P(X < A) = P(Z < (A - mean)/standard deviation)

a) P(notification because amount is very low) = P(X < 2500)

= P(Z < (2500 - 4200)/720)

= P(Z < -2.36)

= 0.0091 = 0.91%

b) P(notification because amount is very high) = P(X > 6000)

= P(Z < (6000 - 4200)/720)

= P(Z < 2.5)

= 1 - P(Z < 2.5)

= 1 - 0.9938

= 0.0062 = 0.62%

c) P(notification) = 0.91% + 0.62% = 1.53%

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.