Question

1.- An expensive watch is powered by a​ 3-volt lithium battery expected to last three years. Suppose the life of the battery has a standard deviation of 0.3 year and is normally distributed.a. Determine the probability that the​ watch's battery will last longer than 3.8 years.b. Calculate the probability that the​ watch's battery will last more than 2.35 years.c. Compute the​ length-of-life value for which 20​% of the​ watch's batteries last longer.

a. The probability that the battery will last longer than 3.8 years is

(Round to four decimal places as​ needed.)

b. The probability that the battery will last more than 2.35 years is

​(Round to four decimal places as​ needed.)

c. The​ length-of-life value for which 20​% of the batteries last longer is : ? years.

​(Round to one decimal place as​ needed.)

2.- For a standardized normal​ distribution, calculate the probabilities below.

a. ​P(z<1.7​)

b. ​P(z≥0.75​)

c. ​P(−1.21<z<1.15​)

a. ​P(z<1.7​)=

​(Round to four decimal places as​ needed.)

b. ​P(z≥0.75​)=

​(Round to four decimal places as​ needed.)

c. ​P(−1.21<z<1.15​)=

​(Round to four decimal places as​ needed.)

Answer 1

Solution: Given that mean U = 3 3 years Standard deviation 5 = 0.3 years P(x > 3.8 years) = 03 |- P(x<3.8) 51- P(Z < 3.8–3 ) = 1-P (ZLOo8 ) {-PZ< 2.6667) 1-(0.9962) 0,3 - PX73,8) (fromz-table) =10.0038 6). P(x>2.35 year) - 03 0.3 - 1- PXC 2.35) 1-.pza 2.35-3) E1- PE=0.65) 1 P.(Z <-201667) 1 - 0.01539 (fromz_table) PCX22035) -10.9846) Let the value for which 20% of the batteries P@> A) = 0.2 PXCA) = 1-012 ce), - 0.8 Ć A-mean ) = 0.8 P(Z < S.D. A-3 = 0.84 (from 2-table 0.3 A-3 = 0.84 0.3 :)

A-3 2 z 0.252 Az 0.252+3 А. 8.252, (rounded one decimal) A=13.3