A population of values has a normal distribution with μ = 202.1 and σ = 24 . You intend to draw a random sample of size n = 64 . Find the probability that a single randomly selected value is between 204.5 and 205.1

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