Question

The owner of an electronic store has read an advertisement from a local competitor that guarantees all high-definition television (HDTV) repairs within four days. Based on his company's experience, he wants to know if he can offer a similar guarantee. His past service records are used to determine the probability distribution shown to the right. # of Days Probability 0.13 1 2 0.24 3 0.24 4 0.25 5 0.14 a. The meanis 3.03 days. (Type the entire number, do not round.) b. The standard deviation is days. (Round to four decimal places as needed.) c. Based on the calculations in parts a and b and the distribution, can the manager offer the same warranty? He could only offer it if there is a 80% probability or less that he won't violate the warranty. Write Yes or No.

Answer 1

SOL!! From given data a²p(a) (x-M) 2 (X-M?)) 2 P(x) 411209 015357 0.13 0:13 0.13 - 0.2546 1-0609 0.96 o.ug 2 0:24 0.000216 2.16 0.0009 3 0.24 0.72 0.2351 0.9409 4 ५ 0-25 3.8809 0.5433 5 0:14 0.7 3.5 10.75 10.6045 1.5689 Total= 1 M = 3.03 a) Mean 3.03 b) Variance = 1.5689 2 = 1.5689 1.5689 Standard deviation (o)= 1.2525 c) P (*>4) = P(X=5) -0.14