Find the probability that a single randomly selected value is between 110.1 and 111.2. P(110.1 < X < 111.2) =

Question

A population of values has a normal distribution with mean=111.2 and standard deviation=5. You intend to draw a random sample of size n=19.

Find the probability that a single randomly selected value is between 110.1 and 111.2.

P(110.1 < X < 111.2) = 

Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

Hi Kaitlyn,

I think they pulled a trick on us with this problem. I wasn’t looking at the question carefully enough—it wants probability of a single value, not a sample mean, which I initially missed. Sorry about that. Anyway, this means you don’t have to account for standard error, so sample size is extraneous.

We can proceed with our old friend z=(x-mu)/sigma. Since we have two values here, I will use Z₁, Z₂, X_1, and X_2.

Z₁=(x₁-mu)/sigma

x₁=110.1

mu=111.2

sigma=5

Z₁=(110.1-111.2)/5

Z₁= -0.22

Now, let’s find Z_2. This will be quite simple because:

X₂=111.2

mu=111.2

sigma=5

But sigma doesn’t matter. The calculation is:

Z₂=(111.2-111.2)/sigma

Z₂=0

Now, for the probability of a value between 110.1 and 111.2

P(-0.22<Z<0)=P(110.1<Z<111.2)

From z-table,

P(Z=0)=0.5

P(Z= -0.22)=0.4129

Do the subtraction

0.5-0.4129=

P=0.0871

Sorry for the confusion. That single value aspect threw me for a loop.