A question on a test is answered correctly by 22% of the students find the probability that in the next 150 students there will be more than 40 correct answers use the normal distribution
To solve this problem, we will use the normal distribution to approximate the probability. Here are the step-by-step calculations:
1. Find the mean (μ) and standard deviation (σ) of the distribution:
Mean (μ) = 150 * 0.22 = 33
Standard Deviation (σ) = sqrt(150 * 0.22 * (1 - 0.22)) = 5.29
2. Convert the problem into a standard normal distribution by calculating the z-score:
z = (x - μ) / σ
where x is the number of correct answers (40), μ is the mean (33), and σ is the standard deviation (5.29)
z = (40 - 33) / 5.29 = 1.32
3. Calculate the probability of getting more than 40 correct answers:
P(X > 40) = 1 - P(X ≤ 40)
P(X > 40) = 1 - P(Z ≤ 1.32)
You can look up the value of P(Z ≤ 1.32) in the standard normal distribution table or use a calculator/statistical software.
The probability P(X > 40) is equivalent to 1 minus the cumulative probability of 1.32.
Let's assume P(Z ≤ 1.32) = 0.9066.
P(X > 40) ≈ 1 - 0.9066 = 0.0934 (approximately)
Therefore, the probability that in the next 150 students there will be more than 40 correct answers is approximately 0.0934 or 9.34%.
To find the probability that in the next 150 students there will be more than 40 correct answers, we can use the normal distribution. Follow these steps to calculate it:
Step 1: Convert the percentage to a decimal.
The percentage of students who answer the question correctly is 22%. Convert it to a decimal by dividing by 100: 22%/100 = 0.22.
Step 2: Find the mean and standard deviation.
The mean of a binomial distribution is given by p * n, where p is the probability of success (0.22) and n is the number of trials (150). So, the mean (μ) = 0.22 * 150 = 33.
The standard deviation (σ) of a binomial distribution is given by the square root of (p * (1 - p) * n). So, the standard deviation (σ) = √(0.22 * (1 - 0.22) * 150) ≈ 4.8.
Step 3: Convert the problem to a standard normal distribution.
To use the normal distribution table, we need to convert the problem to a standard normal distribution by calculating the z-score. The z-score measures how many standard deviations a given value is from the mean and helps us find the corresponding probability.
To find the z-score, use the formula: z = (x - μ) / σ
In this case, x represents the number of correct answers, which is 40.
So, z = (40 - 33) / 4.8 ≈ 1.46
Step 4: Find the probability using the standard normal distribution table.
Using the standard normal distribution table or a calculator, find the probability associated with the z-score of 1.46. The table will give you the probability of values less than the z-score. However, we are interested in finding the probability of more than 40 correct answers.
To find this probability, subtract the cumulative probability of 1.46 (which represents values less than or equal to 40) from 1 (the total probability). P(Z > 1.46) = 1 - P(Z ≤ 1.46).
Look up the probability of 1.46 in the standard normal distribution table (found in most statistics textbooks or online) and subtract it from 1.
Let's assume the probability of 1.46 is approximately 0.9292. So, P(Z > 1.46) = 1 - 0.9292 = 0.0708.
Therefore, the probability that in the next 150 students there will be more than 40 correct answers is approximately 0.0708 or 7.08%.