1.) T or F. The area under the normal curve is split 50/50 about the y-axis. 2.) T or F. A z number of 2.34 means that the data is 2.34 standard deviations below the mean. 30. T or F. the probability to the right of z=-1.34 .0901 4.) T or F. under normal curve, the area to the left of z=0.84 is 0.2
a)
Right
Since mean of standard normal distribution is 0 and P(Z < 0) = P(Z > 0) = 0.5
b)
False
If z = 2.34 (positive value)
This means the data is 2.34 standard deviations above the mean
c)
P(z > -1.34) = P(Z < 1.34) = 0.9099
False
d)
P(z < 0.84) = 0.7995
False.
1.) T or F. The area under the normal curve is split 50/50 about the y-axis....
1. a) About ____ % of the area under the curve of the standard normal distribution is between z=−0.409z=-0.409 and z=0.409z=0.409 (or within 0.409 standard deviations of the mean). b) About ____ % of the area under the curve of the standard normal distribution is outside the interval z=[−0.78,0.78]z=[-0.78,0.78] (or beyond 0.78 standard deviations of the mean). c) About ____ % of the area under the curve of the standard normal distribution is outside the interval z=−0.86z=-0.86 and z=0.86z=0.86 (or...
Answer the following questions about the Standard Normal Curve: a.) Find the area under the Standard Normal curve to the left of z = 1.24 (use 4 decimal places) b.) Find the area under the Standard Normal curve to the right of z = -2.13 (use 4 decimal places) c.) Find the z-value that has 87.7% of the total area under the Standard Normal curve lying to the left of it. d.) Find the z-value that has 20.9% of the total area...
Answer the following questions about the Standard Normal Curve: a.) Find the area under the Standard Normal curve to the left of z = 1.24 (use 4 decimal places) b.) Find the area under the Standard Normal curve to the right of z = -2.13 (use 4 decimal places) c.) Find the z-value that has 87.7% of the total area under the Standard Normal curve lying to the left of it. d.) Find the z-value that has 20.9% of the...
About b6 of the area under the curve of the standard normal distribution is between 0.174 and z = 0.174 (or within 0.174 standard deviations of the mean). z = > Next Question arch
(1 point) Find the Z-score such that: (a) The area under the standard normal curve to its left is 0.8369 Z= (b) The area under the standard normal curve to its left is 0.78 Z= (C) The area under the standard normal curve to its right is 0.1217 Z= (d) The area under the standard normal curve to its right is 0.1206 Z=
To FOUR DECIMAL PLACES: Determine the area under the standard normal curve that lies to the left of Z = –1.31 to the right of Z = –2.47 between Z = –2.47 and Z = –1.31 between Z = 1.31 and Z = 2.47 Find the z-scores that separate the middle 84% of the standard normal distribution from the area in the tails. Find z0.18 a. Find the Z-score corresponding to the 72nd percentile. In other words, find the Z-score...
Find the z-score such that the area under the standard normal curve to the left is 0.98. _______ is the z-score such that the area under the curve to the left is 0.98.Find the z-score such that the area under the standard normal curve to the right is 0.26.The approximate z-score that corresponds to a right tail area of 0.26 is _______
Find the indicated area Under the standard normal curve. To the right of z=2.27 The area to the right of z=2.27 under the standard normal curve is____________. Area under the standard normal distribution to the left of z.(page 1)
Find the indicated area under the standard normal curve. To the left of z = -1.39 and to the right of z = 1.39. The total area to the left of z= -1.39 and to the right of z = 1.39 under the standard normal curve is _______
This Question: 1 pt 33 of 39 (22 complete) the total area under the standard normal curve in parts (a) through (e) below (a) Find the area under the normal curve to the left of z -1 plus the area under the normal curve to the right of z-1 The combined area is Round to four decimal places as needed.) (b) Find the area under the normal curve to the left of z-1.58 plus the area under the normal curve...