Question

QUESTION 3 Find the z value that corresponds to the given area. 0.2157 0 Z 0.0854 QUESTION 4 Find the z value that corresponds to the given area. 0.0721 0.4713 QUESTION 5

Answer 1

0.3) The Nolmal Distribution curve is B 40.2157 1 x OZ to - We know that total area under Nolmal - Curve as 1. and plocł'<t&)0.5 p(-dcx'co) = 0.5 Given the area b/w o and 2 = 0.2157 qe, plo<Z'< z) = 0·2157 (Note:- Hete 7'- standard volmal variate Zzowe finding value] and p(-xx 2x0) + plochez) - 0.5 +0.2157 = 0.7157 (which ºg the Cumulative value] Here pl-dak'<x) = 04157 and plodcz'20.57)=0.7157 l from the Nolmal distribution table the Alea b/w Ed to 0.57) is 0.1157] Hence 750:57

70.0721 - o 7 to 1. * Total value = +1 " The value from na to z is 0.9279 [ value from na to td is + 1 - then value from a to z os 1-0.072) -0.9279 and pl-dez'cz) =0.9219 P (« <z!<1.46)=0-9279 [from dolmal distribution table the value blw Fa to 1.46 is 0.9279 Hence (Z=1.46 O outside the . -2 -0.6 O 1.8 td They asked about the value tie interval fo.6 and 1.8)

1-es, 1-{p(-0.6 <z'<1.80] continuation to a No:5] ® El-[plocz'<0.6) + plocz'<l«8) Note :- Here we take 0.6 instead of o.6. Because of the Area blu o and ta is 0.5 C and ud and o is 0.5 ] - 1-[0.22515 +0.46401) = 0.31018 Atea die outside Ze-06 and 1.8 98 0:31018 .) P(-dazz z)=0:13 Since Cummulative value is less than 0.5 Hence the x value os negative P(-2222-2) = 0.13 P(-zzzco) = PEOK ILO) -0.13 = 0.5-0.13 Plazez'<o) = 0.37 and the cumulative value 93 0-87 (0-5+:37) ::P(1:12 <='ka) = 0.87 :-17=-1:12

0.7) Hond no lontaire : Q.1) Given H=9,=5 Pl@cx<b) =0.34 *ple qez'bog) -0:34 After converting it into Standard Völmal föln brom the volmal distribution table. 1. The value 0.34 is lies blw.o and I @= 9 b b-q=5 -9=5 [b=ly . . The values are 9 and 14. Q-8) Given, H=150, =6 P(x2441) =P ( X5H < 741.750) =P[<<-0.5) (We know the value of shaded Area] -2 -0.5 o . by using Nolmal table. k . Kos * =pl-dez'<o) - P (='<0.5) to = 0.5-0.1915 (P(x<747) = 0.3085