Normal Approximation Help: 200 Student Survey

  • Context: Undergrad 
  • Thread starter Thread starter Ceresu
  • Start date Start date
  • Tags Tags
    Approximation Normal
Click For Summary
SUMMARY

The discussion focuses on using normal approximation to the binomial distribution for estimating the probability of selecting at least 110 girls from a random sample of 200 students, where boys and girls are equally likely to be chosen. The expected number of girls is calculated as 100, with a standard deviation of 5√2. The probability is determined using the z-score formula, specifically calculating the z-score for 109.5 to account for continuity correction. Additionally, it is noted that the binomial distribution is an approximation due to the changing probabilities when selecting without replacement.

PREREQUISITES
  • Understanding of binomial distribution and its properties
  • Knowledge of normal distribution and z-scores
  • Familiarity with continuity correction in statistical calculations
  • Basic grasp of probability theory and expected values
NEXT STEPS
  • Study the application of the Central Limit Theorem in binomial distributions
  • Learn about continuity correction in statistical approximations
  • Explore the differences between binomial and hypergeometric distributions
  • Practice calculating probabilities using normal approximation with various sample sizes
USEFUL FOR

Students in statistics courses, educators teaching probability theory, and anyone interested in understanding statistical approximations in survey sampling.

Ceresu
Messages
1
Reaction score
0
Hello ~

I be in dire need of help with this problem because I fell asleep in math class. Could anyone be so kind as to thoroughly guide me through the following problem?

"A school has enrolled the same number of boys and girls. Two hundred students are selected at random to participate in a survey. Use a normal approximation to the binomial distribution to estimate the probability that at least 110 girls will be selected."

Thanks!
-Mathematically-challenged student :eek:
 
Physics news on Phys.org
If you are selecting 200 students and boys and girls both have probability 1/2 of being selected then the expected number of girls is (1/2)(200)= 100 (in general, with a binomial distribution with probabilities p and 1-p and number n, the expected value is np). The standard deviation is
[itex]\sqrt{200(\frac{1}{2})(\frac{1}{2})}= 5\sqrt{2}[/itex] (in general, it is
[itex]\sqrt{np(1-p)}[/itex]).
"At least 110 girls" means, since the normal distribution does not apply only to integers, "greater than or equal to 109.5" (that's the "half integer correction"). Use a table of normal distribution values to determine the probability that the standard "z" score is greater than [itex]\frac{109.5- 100}{5\sqrt{2}}=1.34[/itex].

(edited thanks to pizzasky)
 
Last edited by a moderator:
Reply

If I am not mistaken, the standard deviation used in this question should be [itex]\sqrt{np(1-p)}=\sqrt{200(\frac{1}{2})(\frac{1}{2})}[/itex].

Apart from that, it would also be helpful to note that the Binomial distribution used in the question IS actually an approximation in itself. After all, we assume constant probability of success when using the Binomial distribution, but this is not really the case here. In choosing the survey participants, we do it without replacement, so the probability of choosing a girl actually changes as more people are chosen. Hence, an important assumption we need to make is that there is a LARGE number of students (both boys and girls) in the school, so the probability of choosing a girl does not change too greatly throughout the whole choosing process.

Hope you get what I mean!:biggrin:
 
Last edited:

Similar threads

Replies
1
Views
2K
Replies
4
Views
4K
  • · Replies 6 ·
Replies
6
Views
12K
  • · Replies 13 ·
Replies
13
Views
4K
  • Poll Poll
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 25 ·
Replies
25
Views
8K
  • · Replies 67 ·
3
Replies
67
Views
17K
  • · Replies 10 ·
Replies
10
Views
2K