Copy and paste your R command and output. <- PLEASE DO THIS!! I don\'t know how
ID: 3046016 • Letter: C
Question
Copy and paste your R command and output. <- PLEASE DO THIS!! I don't know how to do this part!!
Empirical rule Verify the empirical rule by using Table A, software, or a calculator to show that for a normal distribu- tion, the probability (rounded to two decimal places) within
These are the answers, but I need to know how to solve this in like a R CODE
a)1 standard deviation of the mean equals 0.68.
a) The cumulative probability that an observation is falling below the point + is 0.8413. Therefore, the probability that an observation is lying between - and + is 0.68 (2 (0.84 13-0.5) = 2x 0.3413 = 0.6826). That is, for a normal distribution, about 68% observations fall within one standard deviation from the mean. Comment Step 3 of 9 b) The cumulative probability that an observation is falling below the point +20 is 0.9772. Therefore, the probability that an observation is lying between -2 and +2 is 0.95 (2 (0.9772-0.5)-2x 0.4772 = 0.9544). That is, for a normal distribution, about 95% observations fall within two standard deviations from the mean. Comment Step 4 of 9 c) The cumulative probability that an observation is falling below the point +3 is 0.9987. Therefore, the probability that an observation is lying between -3 and +3 is 1.00 (2 (0.9987-0.5)-2 x 0.4987 = 0.9974). That is, for a normal distribution, about 99.7% observations fall within three standard deviations from the mean.Explanation / Answer
Ans:
Use normcdf for Ti-83 or pnorm for R command.
a)For,one standard deviation below or above the mean,z value will be +/-1
We have to check,
P(-1<z<1)=P(z<1)-P(z<-1)=0.8413-0.1587=0.6826 or 68.26%
So,approximately 68%
use normalcdf(-1,1)=0.6826
Using R -command:
pnorm(x) gives the probability that a random value is less than x.
So,use command:
pnorm(1, mean=0, sd=1, lower.tail=TRUE) - pnorm(-1, mean=0, sd=1, lower.tail=TRUE) =0.6826
b)similarly here z=+/-2
pnorm(2, mean=0, sd=1, lower.tail=TRUE) - pnorm(-2, mean=0, sd=1, lower.tail=TRUE)=0.9545
c)z=+/-3
pnorm(3, mean=0, sd=1, lower.tail=TRUE) - pnorm(-3, mean=0, sd=1, lower.tail=TRUE)=0.9973