Question

Suppose X is a Normal random variable with with expected value 16 and standard deviation 1.05. We take a random sample of size n from the distribution of X. Let X be the sample mean. Use R to determine the following: a) Find the probability P(X>16.4) b) Find the probability P(X >16.4) when n 9 c) Find the probability P(X>16.4) when n = 36| d) What is the probability P(15.6 <X <16.4) when n 36? e) What is the standard deviation of X when n = 34 f) What is the probability that X1ı +X2 ... +X20 < 329 g) Copy your R script for the above into the text box here.

Answer 1

We have given that. YNN (4=16,0? (05)) Using R-software srau = 16 ; siqma = 1.05 We have to find P(X>16.4) command using R- software is, > l-proym (16.4.,mu = 16 Sigma = 1.05) (10.3516193 o m |::.pl X > 16:4) =0.35)6193 T (6) We have to find :p ( X > 16:4) when n=9 XNN 11102) = N N (40?/n). ..P1 >16.4) < p / g -4 16.4-ul... Using centra colun olun . llimit theorem | where INNCO.) ePlz 16.4-16 1.05/15 > 1.1429 Llommand using R-software iss >-onorm (1.1429,0,0) C) 6.1265401 Lip C5 > 16.4) when nog is Slanned with damScanner 0.126 540)

o I we have to find PC 1 216.4) wbpo n=36: 62 FN(4,olun AP(F>16.4) = 1/8-4 16.4-4) ननल Using central limit theorem where, ZnNi 11.05 136) = P(Z > 16.4-16 = P(Z > 2.2857) command using R-software ist |>1-prown ( 2.2857,0,1) 51] 0.0111359) --- P6216-4) when 0=36 ic_0:0))]359|| (a) We have to find PC 15.6 < <16.4) when 0=36 XNN(-4,02) XUN( 4,0? 1n) L (15 -6 < x < 6 ) = P ( 15-6 -44 X - L64 - 84 = //15.6-16 ,2 c16.4-16.) 36 10536 -_ Where, ZNN (021) = P(2.285752 < 2.2857) - = (x 2.2.8 %) - P(XK- .2897) - IR-command is |>pnorm (2.2857,0,1)-phorm (-2.2857,0,1) CamScanner

[1] 0.9777282 ...p(15.6 < <16.4) = 0.977282'. Te We have to 1=34 find standard deviation of t when We know that if X NN(4,02) then Ť NN (4,82/n) Here, U = 16, 8²= (1.052 standard deviation of ř=0 Using R-command:_ > sigma = 1:05: n = 34 > sd = sigma / (sqrt (n)) >sd [1] 0:1800735 | ::: Standard deviation of 5 e 0.1800735 (f) | We have to find P ( X1 + X2 + ... + X20 < 329) lie Plexi (329) * Dividing by , =P(X (16.45) =P ( X L 1648-22 canned with I am Scanner

where, INN(0,1) I plz 16.45-16 T 1105 I=plzel.9166) Using R commands >pnorm (1.9166,0,1) [i]'0.9723556 carsparneš nth X 2 +--+X20 (329) 20.9723556 amScanner
The answer to the question (g) in R script is as follows:

The R script is:

> mu=16;sigma=1.05;n=20
> a=((329/20)-mu)/(sigma/sqrt(n))
> p=pnorm(a,0,1)
> p
[1] 0.9723575