Answer:
Step-by-step explanation:
(a)
The distribution of X is Normal Distribution with mean
and Variance 
(b)
The distribution of
is Normal Distribution with mean
and Variance =
.i.e., 
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)

= 1.0328
To find 
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,

So,
Answer is:
0.7605
(e)
Correct option:
No
because Population SD is provided.
(f)
(i)
Q1 is given by:

So,
X = 17 - (0.6745 * 1.0328) = 17 - 0.6966 = 16.3034
So,
Q1 = 16.3034
(ii)
Q3 is given by:

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