Answer:
Step-by-step explanation:
(a)
The distribution of X is Normal Distribution with mean
and Variance ![= \sigma^{2} = 16 \ i.e., X \sim N (17, 16),](https://tex.z-dn.net/?f=%3D%20%5Csigma%5E%7B2%7D%20%3D%2016%20%5C%20i.e.%2C%20X%20%5Csim%20N%20%2817%2C%2016%29%2C)
(b)
The distribution of
is Normal Distribution with mean
and Variance =
.i.e., ![\bar{x}\sim N(17,1.0667)](https://tex.z-dn.net/?f=%5Cbar%7Bx%7D%5Csim%20N%2817%2C1.0667%29)
c)
To find P(15.5 < X < 18):
Case 1: For X from 15.5 to mid value:
Z = (15.5 - 17)/4 = - 0.375
Table of Area Under Standard Normal Curve gives area = 0.1480
Case 2: For X from mid value to 18:
Z = (18 - 17)/4 = 0.25
Table of Area Under Standard Normal Curve gives area = 0.0987
So,
P(15.5 < X< 18) = 0.1480 +0.0987 = 0.2467
So,
Answer is:
0.2467
(d)
![SE = \sigma/\sqrt{n}\\\\= 4/\sqrt{15}](https://tex.z-dn.net/?f=SE%20%3D%20%5Csigma%2F%5Csqrt%7Bn%7D%5C%5C%5C%5C%3D%204%2F%5Csqrt%7B15%7D)
= 1.0328
To find ![P(15.5 < \bar{x}< 18):](https://tex.z-dn.net/?f=P%2815.5%20%3C%20%5Cbar%7Bx%7D%3C%2018%29%3A)
Case 1: For
from 15.5 to mid value:
Z = (15.5 - 17)/1.0328 = - 1.4524
Table of Area Under Standard Normal Curve gives area = 0.4265
Case 2: For X from mid value to 18:
Z = (18 - 17)/1.0328 = 0.9682
Table of Area Under Standard Normal Curve gives area = 0.3340
So,
![P(15.5 < \bar{x}< 18) = 0.4265 + 0.3340 = 0.7605](https://tex.z-dn.net/?f=P%2815.5%20%3C%20%5Cbar%7Bx%7D%3C%2018%29%20%3D%200.4265%20%2B%200.3340%20%3D%200.7605)
So,
Answer is:
0.7605
(e)
Correct option:
No
because Population SD is provided.
(f)
(i)
Q1 is given by:
![- 0.6745 = (\bar{x} - 17)/1.0328](https://tex.z-dn.net/?f=-%200.6745%20%3D%20%28%5Cbar%7Bx%7D%20-%2017%29%2F1.0328)
So,
X = 17 - (0.6745 * 1.0328) = 17 - 0.6966 = 16.3034
So,
Q1 = 16.3034
(ii)
Q3 is given by:
![0.6745 = (\bar{x} - 17)/1.0328](https://tex.z-dn.net/?f=0.6745%20%3D%20%28%5Cbar%7Bx%7D%20-%2017%29%2F1.0328)
So,
X = 17 + (0.6745 * 1.0328) = 17 + 0.6966 = 17.6966
So,
Q3= 17.6966
(iii)
IQR = Q3 - Q1 = 17.6966 - 16.3034 = 1.3932
So
Answers are:
Q1 = 16.3034 ounces
Q3 = 17.6966 Ounces
IQR = 1.3932 Ounces