Thread: Probability of sample - getting Too large a Z value

1. Probability of sample - getting Too large a Z value

I'm working on this more advanced problem:

Population proportion of college students that attend every class is 0.15
What is the probability that out of 100 students sampled, the proportion of those attending every class is between 0.1 and 0.3?

I thought I knew how to solve this, but I keep getting a Z value greater than 3, which is too big to be listed in the Z table that we use. This is what I did:

Use Infinite population formula for Standard deviation(SD) of P:

SD[of P] = SQRT(P(1-P)/n))

so thats: SQRT(0.15(1-0.15)/100)) which = 0.0358

Then using the Z formula which is:

Z = (X[value your using] - Population Proportion)/ SD[of p]

so thats Z = ((0.3-0.15)/0.0358) = 4.1899

right here I'm getting a Z value of 4+, which is too big to use in the Z tables. However the calculations for the 0.1 area works out ok to a Z value of -1.39, but I' not sure how to find the area of the .3, since the Z value is too big for the table. Any help is greatly appreciated.

Basically My idea to solve it is to just substract the 0.1 area from the 0.3 area, which will give me the area between 0.1 and 0.3 which will be the probability that it is between 0.1 and 0.3

I'm working on this more advanced problem:

Population proportion of college students that attend every class is 0.15
What is the probability that out of 100 students sampled, the proportion of those attending every class is between 0.1 and 0.3?

I thought I knew how to solve this, but I keep getting a Z value greater than 3, which is too big to be listed in the Z table that we use. This is what I did:

Use Infinite population formula for Standard deviation(SD) of P:

SD[of P] = SQRT(P(1-P)/n))

so thats: SQRT(0.15(1-0.15)/100)) which = 0.0358

Then using the Z formula which is:

Z = (X[value your using] - Population Proportion)/ SD[of p]

so thats Z = ((0.3-0.15)/0.0358) = 4.1899

right here I'm getting a Z value of 4+, which is too big to use in the Z tables. However the calculations for the 0.1 area works out ok to a Z value of -1.39, but I' not sure how to find the area of the .3, since the Z value is too big for the table. Any help is greatly appreciated.

Basically My idea to solve it is to just substract the 0.1 area from the 0.3 area, which will give me the area between 0.1 and 0.3 which will be the probability that it is between 0.1 and 0.3
If you assume a large number of college students then you can treat it as a sampling with replacement problem. In which case you could re-word the question as:

If the probability that a college student attends every class is 0.15, find the probability that out of 100 students sampled, the number of those attending every class is between 10 and 30.

I'm working on this more advanced problem:

Population proportion of college students that attend every class is 0.15
What is the probability that out of 100 students sampled, the proportion of those attending every class is between 0.1 and 0.3?

I thought I knew how to solve this, but I keep getting a Z value greater than 3, which is too big to be listed in the Z table that we use. This is what I did:

Use Infinite population formula for Standard deviation(SD) of P:

SD[of P] = SQRT(P(1-P)/n))

so thats: SQRT(0.15(1-0.15)/100)) which = 0.0358

Then using the Z formula which is:

Z = (X[value your using] - Population Proportion)/ SD[of p]

so thats Z = ((0.3-0.15)/0.0358) = 4.1899

right here I'm getting a Z value of 4+, which is too big to use in the Z tables. However the calculations for the 0.1 area works out ok to a Z value of -1.39, but I' not sure how to find the area of the .3, since the Z value is too big for the table. Any help is greatly appreciated.

Basically My idea to solve it is to just substract the 0.1 area from the 0.3 area, which will give me the area between 0.1 and 0.3 which will be the probability that it is between 0.1 and 0.3

Don't worry that the .3 transform to a z-score that isn't on the tables.
Just find the probability of exceeding the lower z-score here.

$P(Z>-1.39)$ HOWEVER you can make this more accurate by using the +-.5 correction factor.
This will improve on your estimate of the true probability since we are overlapping a continuous (normal) random variable
over a discrete (binomial) random variable.