# Statistics Question Math

Sam is a customer service officer working in a telecommunication company. According to his experience, time he uses to answer enquiries by email in a day follows a normal distribution with mean of 2.4 hours and standard deviation of 0.8 hour, while time he uses to answer enquiries by mail in a day follows a normal distribution with mean of 3.2 hours and standard deviation of 0.4 hour. Suppose the time he spent on one jobs is independent with the other job.

a. What is the probability that he uses more than 3 hours to answer enquiries by email in a day?

b. What is the probability that he uses more than 6 hours to finish these two jobs in a day?

c. Suppose there is a 79.1% chance that he uses less than T hour to answer enquiries by mail in a day. What is the value of T?

d. Sam works 5 days in a week. What is the probability that there is exactly 4 days he uses more than 3 hours to answer enquires by email in a week?

### 1 個解答

• 6 年前
最愛解答

Let X (in hours) be the time Sam uses to answer enquiries by email in a day, then X ~ N(2.4, 0.8²).

Let Y (in hours) be the time Sam uses to answer enquiries by mail in a day, then Y ~ N(3.2, 0.4²).

(a)

Pr( X > 3 )

= Pr( Z > (3 - 2.4)/ 0.8 )

= Pr( Z > 0.75 )

= 1 - 0.7734

= 0.2266

(b)

Since X and Y are independent, X + Y ~ N(2.4 + 3.2, 0.8² + 0.4²)

Pr( X + Y > 6 )

= Pr( Z > (6 - 2.4 - 3.2)/√(0.8² + 0.4²) )

= Pr( Z > 0.4/√0.8 )

= Pr( Z > 0.447213595 )

≈ Pr( Z > 0.45 )

= 1 - 0.6736

= 0.3264

(c)

Pr( Y < T ) = 0.791

Pr( Z < (T - 3.2)/0.4 ) = 0.791

(T - 3.2)/0.4 = 0.81

T = 3.524

(d) [問句最後的 in a week 應該是 in a day, 反而 exactly 4 days 是指 exactly 4 days in a week 才對。]

We need to use the result of (a).

Let S be the number of days in a week Sam uses more than 3 hours to answer enquires by email in a day, then S ~ B(5, 0.2266)

The required probability is

Pr(S = 4)

= ₅C₄ (0.2266)⁴ (1 - 0.2266)

= 5 (0.2266)⁴ (0.7734)

= 0.010195624