Math Problem Statement

In a recent​ year, the scores for the reading portion of a test were normally​ distributed, with a mean of 21.5 and a standard deviation of 5.3. Complete parts​ (a) through​ (d) below. Question content area bottom Part 1 ​(a) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is less than 17. The probability of a student scoring less than 17 is    ​(Round to four decimal places as​ needed.) Part 2 ​(b) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is between 14.1 and 28.9. The probability of a student scoring between 14.1 and 28.9 is   

​(Round to four decimal places as​ needed.)

Solution

We are working with a normal distribution here, where the mean (μ) is 21.5 and the standard deviation (σ) is 5.3. To solve these parts, we'll use the Z-score formula to convert the raw scores into Z-scores, and then we can find the corresponding probabilities using standard normal distribution tables or a calculator:

Z=XμσZ = \frac{X - \mu}{\sigma}

Part (a) - Probability that a score is less than 17:

  1. Calculate the Z-score for X=17X = 17:

Z=1721.55.3=4.55.30.8491Z = \frac{17 - 21.5}{5.3} = \frac{-4.5}{5.3} \approx -0.8491

  1. Find the probability corresponding to this Z-score. Using the standard normal distribution table or a calculator, the probability for Z=0.8491Z = -0.8491 is approximately:

P(Z<0.8491)0.1985P(Z < -0.8491) \approx 0.1985

Thus, the probability that a randomly selected student scores less than 17 is approximately 0.1985.


Part (b) - Probability that a score is between 14.1 and 28.9:

  1. Calculate the Z-score for 14.1:

Z=14.121.55.3=7.45.31.3962Z = \frac{14.1 - 21.5}{5.3} = \frac{-7.4}{5.3} \approx -1.3962

  1. Calculate the Z-score for 28.9:

Z=28.921.55.3=7.45.31.3962Z = \frac{28.9 - 21.5}{5.3} = \frac{7.4}{5.3} \approx 1.3962

  1. Find the probabilities for these Z-scores:

    • P(Z<1.3962)0.0816P(Z < -1.3962) \approx 0.0816
    • P(Z<1.3962)0.9184P(Z < 1.3962) \approx 0.9184
  2. Subtract the two probabilities to find the probability that the score is between 14.1 and 28.9:

P(14.1<X<28.9)=P(Z<1.3962)P(Z<1.3962)=0.91840.0816=0.8368P(14.1 < X < 28.9) = P(Z < 1.3962) - P(Z < -1.3962) = 0.9184 - 0.0816 = 0.8368

Thus, the probability that a randomly selected student scores between 14.1 and 28.9 is approximately 0.8368.


Would you like more details on this solution or further clarification?

Related Questions:

  1. What is the probability of scoring above 25 on the test?
  2. How would you find the probability for a score less than 10?
  3. What Z-score corresponds to the top 10% of the distribution?
  4. If a student scores 30, what is their percentile rank?
  5. What is the probability of a score between 20 and 22?

Tip:

For normal distributions, knowing how to use Z-scores is crucial for finding probabilities and percentiles, especially when data doesn't match typical values from tables.

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Z-scores
Probability

Formulas

Z = (X - μ) / σ

Theorems

Standard Normal Distribution

Suitable Grade Level

Grades 10-12