Homework: 7.2 Homework 120, score: 0 of 1 pt 21 of22(19complete) Hw score: 61.06

Question

Homework: 7.2 Homework 120, score: 0 of 1 pt 21 of22(19complete) Hw score: 61.06%, 13.43 of 22 pts Hormel × 7.251 Question Help The time required for an automotive center to complete an oll change service on an automobile approximately follows a normal distnbution, with a mean of 19 minudes and a standard deviation of 4 minutes (a) The hall-price. What percent of customers receive the (b) " the automotive center does not want to givethe discount to moe than 2% ofts customers, how long should it make the guararhedtne lim? 10:39 PM (a) T (Round to two decimal places as needed.) Clear All 3 8 6 2

Explanation / Answer

Solution:- Given that mean =19 ,sd = 4
a) P(X > 20) = P(Z > (20-19)/4)
= P(Z > 0.25)
= 1 P(Z < 0.25)
= 10.5987
= 0.4013
= 0.40(rounded)

b) P(X > x) = 0.02
P(Z > (x-19)/4) = 0.02
(x-19)/4 = 2.0537
X = 19 + (2.0537*4)
x = 27.2148
x = 27

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

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