SOLUTION: In a recent​ year, grade 8 Colorado State public school students taking a mathematics assessment test had a mean score of 290 with a standard deviation of 37. Assume that the

Algebra ->  Probability-and-statistics -> SOLUTION: In a recent​ year, grade 8 Colorado State public school students taking a mathematics assessment test had a mean score of 290 with a standard deviation of 37. Assume that the      Log On


   

Question 1171493:
In a recent​ year, grade 8 Colorado State public school students taking a mathematics assessment test had a mean score of 290 with a standard deviation of 37. Assume that the scores are normally distributed.
​A) Find the probability that a student had a score higher than 320. Enter your answer as a decimal rounded to 2 places.
​B) Find the probability that a student had a score between 250 and 300. Enter your answer as
a decimal rounded to 2 places.
​C) If 2000 students are randomly​ selected, how many would you expect to have a test score that is less than​ 280? Round your answer to the nearest whole number.
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​D) What is the score that corresponds to the 99th percentile of all​ scores? Round your answer to the nearest whole number.  
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​E) A random sample of 60 students is drawn from this population. What is the probability that the mean test score of this group is greater than​ 300? Enter your answer as a decimal rounded to 2 places
Answer by CPhill(1987) About Me  (Show Source):
You can put this solution on YOUR website!
Let's solve each part of this problem step-by-step.
**Given Information:**
* Mean (μ) = 290
* Standard deviation (σ) = 37
* Scores are normally distributed.
**A) Probability of a score higher than 320:**
1. **Calculate the z-score:**
* z = (x - μ) / σ
* z = (320 - 290) / 37
* z = 30 / 37 ≈ 0.81
2. **Find the probability:**
* Use a z-table or calculator to find P(Z > 0.81).
* P(Z > 0.81) ≈ 0.2090
* Rounded to 2 decimal places: 0.21
**B) Probability of a score between 250 and 300:**
1. **Calculate the z-scores:**
* z1 = (250 - 290) / 37 = -40 / 37 ≈ -1.08
* z2 = (300 - 290) / 37 = 10 / 37 ≈ 0.27
2. **Find the probabilities:**
* P(Z < 0.27) ≈ 0.6064
* P(Z < -1.08) ≈ 0.1401
3. **Calculate the probability between:**
* P(-1.08 < Z < 0.27) = P(Z < 0.27) - P(Z < -1.08)
* = 0.6064 - 0.1401 ≈ 0.4663
* Rounded to 2 decimal places: 0.47
**C) Expected number of students with a score less than 280 (out of 2000):**
1. **Calculate the z-score:**
* z = (280 - 290) / 37 = -10 / 37 ≈ -0.27
2. **Find the probability:**
* P(Z < -0.27) ≈ 0.3936
3. **Calculate the expected number:**
* Expected number = 2000 * 0.3936 ≈ 787.2
* Rounded to the nearest whole number: 787
**D) Score corresponding to the 99th percentile:**
1. **Find the z-score:**
* Use a z-table or calculator to find the z-score corresponding to the 99th percentile (0.99).
* z ≈ 2.33
2. **Calculate the score:**
* x = μ + zσ
* x = 290 + (2.33 * 37)
* x = 290 + 86.21 ≈ 376.21
* Rounded to the nearest whole number: 376
**E) Probability that the mean of a sample of 60 students is greater than 300:**
1. **Calculate the standard error:**
* SE = σ / √n
* SE = 37 / √60 ≈ 4.77
2. **Calculate the z-score for the sample mean:**
* z = (x̄ - μ) / SE
* z = (300 - 290) / 4.77 ≈ 2.10
3. **Find the probability:**
* P(Z > 2.10) ≈ 0.0179
* Rounded to 2 decimal places: 0.02