Question

Exercise 6-23 Algo Let X be normally distributed with mean y = 2,900 and standard deviation o = 1,600. [You may find it useful to reference the z table.) a. Find x such that Pixs x) = 0.9382. (Round "z" value to 2 decimal places, and final answer to nearest whole number.) b. Find x such that PIX> x) = 0.025. (Round "z" value to 2 decimal places, and final answer to nearest whole number.)

Answer 1

Solutions - let X be somallei diftributed coith Meon Elie 2900:- Required information is give by, el= 2900 G-1600 Calculate in @ Find y Cuch that pcx= x= 0.9882:- he hale to obtained "&" value by using given data is here, ->PCX<X)= 0.2382 PCX-< X-RJ J50-9882 spets z)= 0.9832 - Z=t1 [o-9362) Z-1.5398 7 -1.54 X-4, 1-54 x-lis 6*1.54] Xall+cr*1.54) = 2900+ [1600*1.54) = 2900 + 2u64 x = 5364] Therefore, x-alue = 9364]

Find the x Guch that permx)=0.095:- gives chra is We have to obtained the x vale by casing < PC X >x= 0.005 » 1-PCXcXJ=0.005 = Pcxcx= 10.0.25 -Pexex < 0.975 -> PCkone KX) = 0975 3p Czez]=0.975 => 7a 74100 -0757 =) X-G = 1.96 - X-2000 =1-96 1600 = &-2900] = [-a6 * 1600] >> X 240 p = (313b X = 270 + 306 y = 6036 ) Therefore x = 6o36) © Find x such that pc 2200 2x < x] =0.1917 :- to obtained the x value by casing given clate is that have to obtained the

here 1600 >PC 29005 XCX)= 0.1917 -> PCXcnJ5 PCXC9900) = 0.1017 PRX-9 -94]-PX-9 Dae-9J co lại, plzzz)=P[7< 900-2900] =0.1217 >> PC 727)=P[tag] =0.1917 -> PC Zeta0.5 =0.1217 - PCZ< Z= 0.5 to. 1017 »przat) = 0.6017 Z> +1006211] -> X 20.30 aa -> X-lie co.3099*]

- X= ll 10-3099*o) x = 2a007 co.3099*16] = 290074a5.84 FX =6395.84 - 3346) Therefore x= (6396) Find PCXS XI= 0.UBU0, X = 2: - value is the obtained the x he halk to fers

..(1840 >> PCX X-4 -0.41840 - p[Zz 7)=0.4IBUO >> Zatco.41840) Eyje 0-000? (x-4)= CO OUol *o) xzel + Co-0401 +1600) X z 9900+ [64.16) -13964. 16) =

Therefore, x value ir 2964-16 ~ (2964)