Use normal approximation(Z-score for p-hat) to find the probability that student scores 80% or lower on a 100 question quiz.
A)If the test contains 250 questions, what is the probability that student will score 80% or lower
What type of questions are on the test? True-false? Multiple choice?
To find the probability that a student will score 80% or lower on a quiz using a normal approximation, we can first calculate the z-score for the given score.
The z-score represents the number of standard deviations a given value is from the mean. It can be calculated using the formula:
z = (x - μ) / σ
Where:
x is the given score (in this case, 80% of the total number of questions)
μ is the mean
σ is the standard deviation
To calculate the z-score, we need to know the mean and standard deviation. Since the question only provides information about the number of questions, we will assume that the quiz scores follow a normal distribution.
For these types of problems, we can use the continuity correction formula when dealing with a discrete distribution like the binomial distribution. The continuity correction formula adjusts the boundaries of the probability distribution to account for the fact that continuous distributions have an infinite number of possible values.
To find the z-score, we need to determine the mean (μ) and standard deviation (σ) for the 100-question quiz.
Since the mean of the sample distribution for a binomial distribution is given by μ = np, where n is the number of trials (100 in this case) and p is the probability of success on a single trial (80% or 0.8), we can calculate the mean:
μ = np = 100 * 0.8 = 80
The standard deviation (σ) of a binomial distribution is given by σ = √(np(1-p)). Plugging in the values:
σ = √(100 * 0.8 * (1 - 0.8)) = √(100 * 0.8 * 0.2) = √(16) = 4
Now that we have the mean and standard deviation, we can calculate the z-score for the given score of 80% or lower.
z = (x - μ) / σ = (80 - 80) / 4 = 0 / 4 = 0
Next, we need to find the corresponding probability using the standard normal distribution table or z-table.
The probability that a student will score 80% or lower can be calculated by finding the cumulative probability to the left of the z-score of 0.
Looking up the z-score of 0 in the standard normal distribution table, we find that the corresponding cumulative probability is 0.5000.
Therefore, the probability that a student will score 80% or lower on a 100-question quiz is 0.5000 or 50%.
Note: The normal approximation is valid when both np and n(1-p) are greater than or equal to 5. In this case, np = 100 * 0.8 = 80 and n(1-p) = 100 * (1 - 0.8) = 20, which satisfy the condition.