Learning Activity 1. Evaluate P(-1.25 SZ s 2.21) A. 0.1 192 B. 0.4285 c, 0.8808

Question

Learning Activity 1. Evaluate P(-1.25 SZ s 2.21) A. 0.1 192 B. 0.4285 c, 0.8808 D. 0.9864 2. Letz be a standard normal random variable. Find the value of Z such that 0.67 of the area under the curve lies to the right of Z. A) z = -0.44 B) 7 = 0.44 c) = 0.7486 D) z = 0.6293 Suppose that individuals applying for a driver's license in South Carolina are given 4 attempts to pass the driver's test. The following probability distribution shows the number of attempts, X, that were required by individuals who passed their exam in 2016. PX) 0.25 030 0.15 0.30 9 3, What is the probability that a randomly selected individual who attained a driver's license required more than one attempt? A. 0.25 B. 0.45 c, 0.65 D. 0.75 4 What is the expected number of attempts? A 2.5 attempts B. 3.2 attempts C. 1 attempt D. 2.8 attempts 5. What is the variance of attempts? A . 1.16 attempts B. 1.35 attempts c. 1.82 attempts D. 25 attempt

Explanation / Answer

1) P(-1.25 < Z < 2.21) = P(Z < 2.21) - P(Z < - 1.25)

= 0.9864 - 0.1056

= 0.8808

Option-C) 0.8808

2) Option-B) Z = 0.44

3) P(X > 1) = 1 - P(X = 1) = 1 - 0.25 = 0.75

Option-D) 0.75

4) E(X) = 1 * 0.25 + 2 * 0.3 + 3 * 0.15 + 4 * 0.3 = 2.5

Option-A) 2.5

5) E(X2) = 12 * 0.25 + 22 * 0.3 + 32 * 0.15 + 42 * 0.3 = 7.6

Var(X) = E(X2) - (E(X))2 = 7.6 - 2.52 = 1.35

Option-B) 1.35

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

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