. O 2/18 points 1 Previous Answers DevoreSta19 4 E.506 XP Construct a normal pro

Question

. O 2/18 points 1 Previous Answers DevoreSta19 4 E.506 XP Construct a normal probability plot for the following sample of observations on coating thickness for low-viscosity paint 0.81 0.87 0.90 1.03 109 1.10 1.27 1.30 1.49 1.50 1.61 1.61 1.67 1.72 1.78 1.84 Determine the z percentile associated with each sample observation. (Round your answers to two decimal places.) Sample observation 0.81 0.87 0.90 1.03 1.09 1.10 1.27 1.30 z percentile Sample observation 1.49 1.50 1.61 1.61 1.67 1.72 1.78 1.84

Explanation / Answer

Assuming standard normal distribution,

P(X < 0.81) = 0.791. So 0.81 is 79.1st percentile. (Ans).

P(X < 0.87) = 0.8078. So 0.87 is 80.78-th percentile. (Ans).

P(X < 0.90) = 0.8159. So 0.90 is 81.59-th percentile. (Ans).

P(X < 1.03) = 0.8485. So 1.03 is 84.85-th percentile. (Ans).

P(X < 1.09) = 0.8621. So 1.09 is 86.21st percentile. (Ans).

P(X < 1.10) = 0.8643. So 1.10 is 86.43rd percentile. (Ans).

P(X < 1.27) = 0.898. So 1.27 is 89.8-th percentile. (Ans).

P(X < 1.30) = 0.9032. So, 1.30 is 90.32nd percentile. (Ans).

P(X < 1.49) = 0.9319. So, 1.49 is 93.19-th percentile. (Ans).

P(X < 1.50) = 0.9332. So, 1.50 is 93.32nd percentile. (Ans).

P(X < 1.61) = 0.9463. So, 1.61 is 94.63rd percentile. (Ans).

P(X < 1.67) = 0.9525. So, 1.67 is 95.25-th percentile. (Ans).

P(X < 1.72) = 0.9573. So, 1.72 is 95.73rd percentile. (Ans).

P(X < 1.78) = 0.9625. So, 1.78 is 96.25-th percentile. (Ans).

P(X < 1.84) = 0.9671. So, 1.84 is 96.71st percentile. (Ans).

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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