1. Let Z the standard normal distribution. (a) find z so that the area under the
ID: 2916412 • Letter: 1
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1. Let Z the standard normal distribution. (a) find z so that the area under the density curve and to theleft of z is 0.4266. (b) find z so that area under density curve and the right of zis 0.8051. (d) find z so that P(Z<z)=0.7416. (e) find z so that P(-z<Z<z)=0.4573. (f) find z so that P(-1.54<Z<z)=0.7895. p/s help me showworking on how you do it. 1. Let Z the standard normal distribution. (a) find z so that the area under the density curve and to theleft of z is 0.4266. (b) find z so that area under density curve and the right of zis 0.8051. (d) find z so that P(Z<z)=0.7416. (e) find z so that P(-z<Z<z)=0.4573. (f) find z so that P(-1.54<Z<z)=0.7895. p/s help me showworking on how you do it.Explanation / Answer
By using thenormal tables website(http://www.stattrek.com/Tables/Normal.aspx)we have the following results(a) find z so thatthe area under the density curve and to the left of z is0.4266. Z =-0.185 (b) find z so thatarea under density curve and the right of z is0.8051. Z= 0.860 (d) find z so thatP(Z<z)=0.7416. Z=0.648 (e) find z so thatP(-z<Z<z)=0.4573. (z)-(-z)=0.4573. implies(z)-1+(z)=0.4573. implies2(z)-1=0.4573 implies (z)= 0.72865 By using normaltable values we have Z=0.609 By using normaltable values we have Z=0.609 (f) find z so thatP(-1.54<Z<z)=0.7895. p/s help me show working on how you doit. (z)-(-1.54)=0.7895 (z)-1+(1.54)=0.7895 (z)-1+0.93822=0.7895 implies (z) =0.85128 z= 1.042 z= 1.042
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