Find the indicated area under the curve of the standard normal distribution; then convert it to a percentage and fill in the blank.
About ______% of the area is between z equals minus 1 and z equals 1 (or within 1 standard deviation of the mean).
68.2 7%
this area can be find out with the help of normal distribution table or Z table
first of all we will find out area below 1
then we will find out area below -1
after subtracting we get 0.6 827
hence answer is 68.2 7%
Find the indicated area under the curve of the standard normal distribution; then convert it to...
Find the indicated area under the curve of the standard normal distribution, then convert it to a percentage and fill in the blank. About _______ % of the area is between z= - 3 and z = 3 (or within 3 standard deviations of the mean).
Find the indicated area under the curve of the standard normal distribution, then convert it to a percentage and fill in the blank. About _____% of the area is between zequals=negative 1.4−1.4 and zequals=1.41.4 (or within 1.41.4 standard deviationsdeviations of the mean).Click to view page 1 of the table. LOADING... Click to view page 2 of the table. LOADING... About nothing% of the area is between zequals=negative 1.4−1.4 and zequals=1.41.4 (or within 1.41.4 standard deviationsdeviations of the mean). (Round to...
Find the indicated area under the curve of the standard normal distribution, then convert it to a percentage and fill in the blank.About _____% of the area is between z=-2.2 and z=2.2 (or within 2.2 standard deviations of the mean).About ____% of the area is between z=-2.2 and z=2.2 (or within 2.2 standard deviations of the mean).
Using the proper functions on a TI 84 Plus: Find the indicated area under the curve of the standard normal distribution; then convert it to a percentage and fill in the blank. About ____ % of the area is between z = -3 and z = 3
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...
Find the indicated area under the standard normal curve. To the right of zequals 1.16Find the indicated area under the standard normal curve. To the right of z equals 1.16
1) Find the area under the standard normal curve to the right of z= -0.62. Round your answer to four decimal places. 2) Find the following probability for the standard normal distribution. Round your answer to four decimal places. P( z < - 1.85) = 3) Obtain the following probability for the standard normal distribution. P(z<-5.43)= 4) Use a table, calculator, or computer to find the specified area under a standard normal curve. Round your answers to 4 decimal places....
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. Sketch the area under the standard normal curve over the indicated interval and find the specified area. (Round your answer to four decimal places.) The area to the left of z = 0 is _______ 2. Sketch the area under the standard normal curve over the indicated interval and find the specified area. (Round your answer to four decimal places.) The area to the left of z = −0.49 is _______ 3. Sketch the area under the standard normal...
a) Sketch the area under the standard normal curve over the indicated interval and find the specified area. (Round your answer to four decimal places.) The area to the right of z = 1.51 is ____ b) The Customer Service Center in a large New York department store has determined that the amount of time spent with a customer about a complaint is normally distributed, with a mean of 9.5 minutes and a standard deviation of 2.1 minutes. What is...