Question

9. Find the probability P(140 < x < 149). The μ= 149 and σ= 5.18. Round to 4DP.

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10. Find the probability using the standard normal distribution. Round to 4DP.

P(z < −1.35 or z > 1.35)

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For 12 – 15, round to 2DP.

12. Find the z-score for which 70% of the area is to the right. ________

13. Find the z-score that corresponds to P₆. ________

14. Find the z-scores for which 60% of the distribution’s area lies between − z and z.

You may want to sketch a drawing.

________

Answer 1

solution Given that M=149 6=5.18 P(140<x< 149) 140-149 149-149 5-18 5.18 EPL - 9az- 5.18 2018, EP(-1.74 caso) EP (zco) - P(Z <-1024) Using z table = 0.5 0.0409 = 0.459) 10) PC zc-1-35 or 2 >1.35) =p(2<-1035) or P(Z31:35) = 0.0885 -1-P(Z <1:35) - 1 - 0-9115 20.0885 20.0885 +0.0885 = 0.1770

Using standard normal table p(2>2) = 70% =I-P (252)=0.70 =P(Z <2) 51-070 = P(Z(2)=0:30 = P(Z <-0.52) 20:30 2=-052 -0.52 13) P6 .P(242)=6% =P(Z <2)=0.06 EP(ZC-1055) 20.06 2= -1:55 -loss

1) . Using standard normal table P(-2<2<2)=60% Ep(2<3)=P(Z<-2) =0.60 =20(2<2) -1=0.60 =2p (262)=1+0.60 =2p(252)=1.60 =P(Z <2)=160/2 =P(ZCZ) = 0.80 = P(Z <0.84)=0.80 ZI 0.84 p(-0.8442 <0.84)=60%. 20.84 0.84|