The scores of the students on a standardized test are normally distributed, with a mean of 500 and a standard deviation of 110. What is the probability that a randomly selected student has a score between 350 and 550? Use the portion of the standard normal table below to help answer the question.
| z | Probability |
| :----- | :---------- |
| 0.00 | 0.5000 |
| 0.25 | 0.5987 |
| 0.35 | 0.6368 |
| 0.45 | 0.6736 |
| 1.00 | 0.8413 |
| 1.26 | 0.8961 |
| 1.35 | 0.9115 |
| 1.36 | 0.9131 |
A. 9%
B. 24%
C. 59%
Answer (2)
The probability that a randomly selected student scores between 350 and 550 on the test is approximately 58.67%. Based on the provided options, the correct answer is C. 59%. In this calculation, we used z-scores and a standard normal table to find the required probabilities.
;
Calculate the z-score for 350: z 350 = 110 350 − 500 ≈ − 1.36 .
Calculate the z-score for 550: z 550 = 110 550 − 500 ≈ 0.45 .
Find the corresponding probabilities from the table: P ( z < − 1.36 ) = 1 − 0.9131 = 0.0869 and P ( z < 0.45 ) = 0.6736 .
Calculate the probability: P ( 350 < x < 550 ) = 0.6736 − 0.0869 = 0.59 .
Explanation
Understand the problem and provided data We are given a problem where student scores on a standardized test are normally distributed with a mean of 500 and a standard deviation of 110. We need to find the probability that a randomly selected student has a score between 350 and 550, using the provided standard normal table.
Calculate the z-scores First, we need to calculate the z-scores for both 350 and 550 using the formula: z = σ x − μ where x is the score, μ is the mean, and σ is the standard deviation.
Compute z-scores for 350 and 550 For x = 350 :
z 350 = 110 350 − 500 = 110 − 150 ≈ − 1.36 For x = 550 :
z_{550} = \frac{550 - 500}{110} = \frac{50}{110} \approx 0.45 4. Find probabilities from the standard normal table Now, we look up the probabilities corresponding to these z-scores in the standard normal table. For $z_{350} = -1.36$, the table gives us the probability for $z = 1.36$, which is $0.9131$. Since the normal distribution is symmetric, the probability that $z < -1.36$ is $1 - 0.9131 = 0.0869$. For $z_{550} = 0.45$, the table gives us the probability $0.6736$. 5. Calculate the probability between 350 and 550 The probability that a student's score is between 350 and 550 is the difference between the probabilities corresponding to the z-scores of 550 and 350: P(350 < x < 550) = P(z < 0.45) - P(z < -1.36) = 0.6736 - 0.0869 = 0.5867 So, the probability is approximately 0.5867 , or 58.67% .
State the final answer Therefore, the probability that a randomly selected student has a score between 350 and 550 is approximately 59%.
Examples
Understanding normal distributions and probabilities is crucial in many real-world scenarios. For instance, in quality control, manufacturers use these concepts to determine the probability that a product meets certain specifications. Similarly, in finance, analysts use normal distributions to model stock prices and assess the likelihood of different investment outcomes. In education, standardized test scores, like the one in this problem, are often normally distributed, allowing educators to understand the distribution of student performance and identify areas for improvement. By calculating probabilities, we can make informed decisions and predictions in various fields.
