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)
To solve this question, you can use the binomial distribution. The binomial distribution is commonly used to model situations where there are a fixed number of independent trials, and each trial has only two possible outcomes (in this case, whether a student has head lice or not).
The probability mass function for the binomial distribution is given by:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
where:
- n is the total number of trials (in this case, the number of students that were checked by the health nurse, which is 120)
- k is the number of successful trials (in this case, the number of students with head lice)
- (n choose k) is the binomial coefficient, which can be calculated as n! / (k! * (n - k)!)
- p is the probability of success (in this case, the probability that a student has head lice, which is 0.05 or 5%)
To find the probability that exactly 6, 7, or 8 students in the random sample have head lice, you can calculate:
P(X = 6) = (120 choose 6) * (0.05^6) * (0.95^114)
P(X = 7) = (120 choose 7) * (0.05^7) * (0.95^113)
P(X = 8) = (120 choose 8) * (0.05^8) * (0.95^112)
You can calculate these probabilities individually using a combination formula and then multiplying the resulting values with the corresponding probabilities. For example:
P(X = 6) = (120! / (6! * (120 - 6)!)) * (0.05^6) * (0.95^114)
However, this method can be quite tedious and time-consuming, especially for larger numbers. An alternative approach is to use an approximate method and apply the properties of a normal distribution.
Since the mean is determined by the product of the total number of students (120) and the probability of success (0.05), the mean is equal to 6. The standard deviation can be calculated as the square root of (n * p * (1 - p)), which in this case is approximately 2.37.
To find the probability that exactly 6, 7, or 8 students have head lice, you can use the properties of a normal distribution. The probability of X being less than or equal to a given value can be approximated using a cumulative normal distribution table.
For example, to find the probability of exactly 8 students having head lice, you can approximate it as the probability of X being less than or equal to 8.5 (since 8.5 is halfway between 8 and 9). Using the cumulative normal distribution table, you can find the corresponding probability.
Repeat this process for exactly 6 and exactly 7, approximating them as 6.5 and 7.5 respectively.
This approximate method using the properties of a normal distribution can provide a quicker and simpler way to calculate the probabilities compared to the exact combinatorial method, especially if you have access to a cumulative normal distribution table.