Answer: p= 0.0532
Step-by-step explanation: from the question
u= population mean = 202.1
σ= population standard deviation = 24
n = sample size = 64.
Since the sample size is greater than 30 and the population standard deviation is known, then the z score is the appropriate to get the probabilistic value.
Z = (x - u) /(σ/√n)
Where x = sample mean.
At x = 204.5, we have the z score as
Z = 204.5 - 202.1/ (24/√64)
Z = 2.4/ (24/8)
Z = 2.4/ 3
Z = 0.8
At x = 205.1, we have the z score as
Z = 205.1 - 202.1/ (24/√64)
Z = 3/ (24/8)
Z = 3/3
Z = 1.
Since the distribution is normal, we compute the probability value attached to the z score using the z table.
Graphical illustration as attachment in the answer.
Note the area under the z score is the probability value of the z score.
As we can see, the area to the left of the shaded area is a, the shaded area is b and the area to the right of the shaded area is c
We have that
a + b + c = 1
Note, the table I'm using only gives area to the left of the distribution.
Area a
To get area a, we use a z score table whose area is to the left of the distribution.
At 0.8, according to the table the area left of 0.8 is 0.78814.
Area c
Area c is right of the distribution and our table is left of the distribution, but 1 - area to the left = area to the right ( since area to left + area to right =1).
Hence area to the left of 1.0 = 0.84134
Area c = 1 - 0.84134 = 0.15866.
Thus, a = 0.78814, c = 0.15866
a + b + c = 1
0.78814 + b + 0.15866 = 1
0.9468 + b = 1
b = 1 - 0.9468
b = 0.0532.
Hence the probability of random variable between 204.5 & 205.1 is 0.0532
