Mr. Smith has 100 students in his Algebra classes. The grades of a quiz are normally distributed with a mean of 70 and a standard deviation of 4. Sally claims that 75 students scored between 66 and 74 on this quiz. Given that this is normally distributed, is Sally correct? Justify your answer using mathematical reasoning and your knowledge of normal distributions.
this is -+ one stad deviantion. 68 percent sare within one standard deviation.
I need more accurate answers please
To determine whether Sally's claim is correct, we need to calculate the probability that exactly 75 out of 100 students scored between 66 and 74 on the quiz. Since the quiz grades are normally distributed with a mean of 70 and a standard deviation of 4, we can use the properties of the normal distribution to calculate this probability.
First, we need to standardize the range of scores from 66 to 74. To do this, we need to calculate the z-scores for the lower and upper bounds of the range. The z-score formula is given by:
z = (x - μ) / σ
Where:
- x is the value we want to standardize (in this case, the lower and upper bounds of the range)
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
For the lower bound:
z_lower = (66 - 70) / 4 = -1
For the upper bound:
z_upper = (74 - 70) / 4 = 1
Next, we need to find the cumulative probability for both lower and upper z-scores. This represents the probability of getting a score less than or equal to a particular z-score. We can use a standard normal distribution table or a calculator/statistical software to find these probabilities.
Using the standard normal distribution table, we find:
P(z ≤ -1) = 0.1587
P(z ≤ 1) = 0.8413
The probability of getting a score between 66 and 74 can be calculated by subtracting the cumulative probability of the lower bound from the cumulative probability of the upper bound:
P(66 ≤ x ≤ 74) = P(z ≤ 1) - P(z ≤ -1)
= 0.8413 - 0.1587
= 0.6826
This means that approximately 68.26% of students score between 66 and 74 on the quiz.
To determine whether Sally's claim is correct, we need to compare this probability to her claim of 75 students scoring in this range out of 100. If Sally's claim is accurate, the probability of 75 students scoring between 66 and 74 should be reasonably high.
To calculate the probability of exactly 75 students scoring between 66 and 74, we can use the binomial distribution. The formula for calculating binomial probabilities is given by:
P(X = k) = (nCk) * p^k * (1 - p)^(n - k)
Where:
- n is the number of trials (number of students in this case, which is 100)
- k is the number of successful outcomes (number of students scoring between 66 and 74)
- p is the probability of a successful outcome in a single trial (probability obtained earlier, which is approximately 0.6826)
- nCk is the number of combinations of n items taken k at a time
We can substitute these values into the formula:
P(X = 75) = (100C75) * 0.6826^75 * (1 - 0.6826)^(100 - 75)
This calculation involves a large number of combinations and is cumbersome to compute manually. Instead, we can use a calculator or statistical software to evaluate this probability.
After calculating the probability, compare it to a reasonable threshold, such as 0.05, to determine whether Sally's claim is accurate. If the probability is below the threshold, it suggests that Sally's claim is unlikely to be true. Conversely, if the probability is above the threshold, it supports Sally's claim.
So, based on the mathematical reasoning and understanding of the normal distribution, we can evaluate Sally's claim and determine if it is correct.