# percentage of Z-score is Between -2.06 and -1.23
ie P(-2.06<Z<-1.23)
P(-2.06<Z<-1.23)=P(Z<-1.23)-P(Z<-2.06)
=0.1093-0.0197=0.0896
#Value of z is obtain from standard normal table
P(-2.06<Z<-1.23) =0.0896 ie 8.96 %
#percentage of Z-score is Between -2.06 and -1.23 is 8.96 %
Question 9 (1 point) Find the percent of z-scores between z = - 2.06 and z...
Find the percent of the total area under the standard normal curve between the following z-scores. Z= -1.79 and z=1.69 The percent of the total area between Z= -1.79 and z=1.69 is ___%. Round to the nearest hundreth as needed.
find the percent of the total area under the standard normal curve between the following z-scores. z=-1.3 and z=-0.5
find p(1.1<z<2.3 [find the area between these two z scores]
Find the z-scores for which 75% of the distribution's area lies between -z and z.
Find the z-scores for which 88% of the distribution's area lies between minus−z and z.
Find the z-scores for which 40% of the distribution's area lies between -z and z Click to view page 1 of the table. Click (o view page 2 of the table The z-scores are (Use a comma to separate answers as needed. Round to two decimal places as needed.) Enter your answer in the answer box. гу re to search O 89
Find the indicated z-score. Find the z-scores for which 98% of the distribution's area lies between -z and z. (-0.99, 0.99) (-1.645, 1.645) (-1.96, 1.96) (-2.33, 2.33)
#10
Use the standard normal distribution table to determine the percent of data between z=0.15 and z= 1.23. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. The percent of data between z= 0.15 and z= 1.23 is (Round to two decimal places as needed.) %
please answer all question 9-11
Question 9 2 pts z scores are useful because they: o reduce the probability of Type I and Type Il errors, allow permit the transformation of raw scores into percentiles. o transform linear scores into nonlinear scores, convert nonlinear scores back into linear scores, and allow us to obtain comparisons between nonlinear and linear scores. Pelation to the mean of its underlying population allow comparisons to be made between stores from different distributions, and permit...
Use the z-score table to answer the question. Note: Round z-scores to the nearest hundredth and then find the required A values using the table. A psychologist finds that the intelligence quotients of a group of patients are normally distributed, with a mean of 96 and a standard deviation of 16. Find the percent of the patients with the following IQs. (a) above 108 % (b) between 84 and 114 %