# stats

posted by
**Anonymous**
.

Head lice can be a problem in classrooms. From the past experience, suppose it is predicted that 5% of the students will get lice if there are some reported cases. The health nurse checks 120 randomly chosen students and records the number of students that have head lice. What is the probability that exactly 6, 7, 8 students in the random sample have head ice?”

How would I do this question??

I presume you mean that there is a 5% chance that a student, drawn at random, will have lice.

You could solve this exactly as a combinatorial problem. The probability of seeing exactly 6 is:

(.95^114 * .05^6)*Z, where Z is 120!/(6!*114!) (where ! means factorial). Repeat for exactly 7 and exactly 8.

Or you could solve approximately and use the properties of a binominal distribution. Here, the expected mean is .05*120=6, the standard deviation is sqrt(120*.05*95)=2.37 So, 8 is .84 standard deviations away from the mean. Use a cumulative normal distribution table to find your probability. (But, I wouldnt use 8, I would use 8.5)