can someone check these please?
find the area under the standard normal distribution curve.
between z=0 and z=1.89
0=.500
1.89=.9706
.500+.9706= .4706
to the left of z= -.75
.2266
between z= .24 and z= -1.12
.24= .5948
-.1.12= .1314
.5948 - .1314= .4634
find the probability of each, using standard normal distribution:
p(z>.82)
1.000- .7939= .2061
p(1.12<z< 1.43)
.9236-.8686= .055
find the z score-
p(z> .0239)
z= -1.98
thanks!
find the area under the standard normal distribution curve.
between z=0 and z=1.89
0=.500
1.89=.9706
.500+.9706= .4706
to the left of z= -.75
.2266
between z= .24 and z= -1.12
.24= .5948
-.1.12= .1314
.5948 - .1314= .4634
find the probability of each, using standard normal distribution:
p(z>.82)
1.000- .7939= .2061
p(1.12<z< 1.43)
.9236-.8686= .055
find the z score-
p(z> .0239)
z= -1.98
thanks!