Answer)
From z table, p(z<0.52) = 0.6985
We know that total area under normal curve is 1
So, p(z>0.52) = 1 - 0.6985 = 0.3015
Question 35 Using the Standard Normal (Z) Distribution Table, find P (Z> 0.52) O .3015 O...
5. Use the Standard Normal Distribution table to find P (Z<-0.69)
Find: P(-2.36 < Z < -1.04 ) using the standard normal distribution table. O a. 1401 b..0717 C..1583 d. 9066 e. 8417 Of..0934
Find the indicated probability using the standard normal distribution. P(z>2.73) dard normal table
For a standard normal distribution, find: P(-2.43 < z < -1.87) For a standard normal distribution, find: P(-2.43 <z<-1.87) Submit License Question 3. Points possible: 1 This is attempt 1 of 3.
For a standard normal distribution, find: P(Z < -0.02) Submit Question
Question Help The test statistic of z= -2.90 is obtained when testing the claim that p<0.52 a. Using a significance level of a= 0.01, find the critical value(s). b. Should we reject Ho or should we fail to reject Ho? Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. The critical value(s) is/are z = (Round to two decimal places as needed. Use a...
Standard Normal distribution. With regards to a standard normal distribution complete the following: (a) Find P(Z > 0), the proportion of the standard normal distribution above the z-score of 0. (b) Find P(Z <-0.75), the proportion of the standard normal distribution below the Z-score of -0.75 (c) Find P(-1.15<z <2.04). (d) Find P(Z > -1.25). (e) Find the Z-score corresponding to Pso, the 90th percentile value.
Find the indicated probability using the standard normal distribution. Find the indicated probability using the standard normal distribution. P(-0.39<z<0) Click here to view page 1 of the standard normal table. Click here to view page 2 of the standard normal table. P(-0.39<z<0)= (Round to four decimal places as needed.)
For a standard normal distribution, find: P(-1.95<z<0.09)
For a standard normal distribution, find: P(0.61 < z < 2.92)