Math Problem Statement

A normally distributed population has a mean of 575 and a standard deviation of 40. a. Determine the probability that a random sample of size 16 selected from this population will have a sample mean less than 555. b. Determine the probability that a random sample of size 25 selected from the population will have a sample mean greater than or equal to 593. Question content area bottom Part 1 a. Upper P left parenthesis x overbar less than 555 right parenthesisequals

enter your response here ​(Round to four decimal places as​ needed.)

Solution

To solve these problems, we will use the sampling distribution of the sample mean. Since the population is normally distributed, we can apply the following steps for both questions:

Key Formulas:

  1. Standard error of the mean (SE): SE=σnSE = \frac{\sigma}{\sqrt{n}} where:

    • σ\sigma is the population standard deviation,
    • nn is the sample size.
  2. Z-score for the sample mean: Z=xˉμSEZ = \frac{\bar{x} - \mu}{SE} where:

    • xˉ\bar{x} is the sample mean we are interested in,
    • μ\mu is the population mean,
    • SESE is the standard error of the mean.
  3. Probability: The probability corresponds to the area under the normal curve to the left (or right) of the Z-score, which can be looked up in the Z-table.

Part (a): Sample size = 16, Find P(xˉ\bar{x} < 555)

Given:

  • Population mean (μ\mu) = 575
  • Population standard deviation (σ\sigma) = 40
  • Sample size (nn) = 16
  • Desired sample mean (xˉ\bar{x}) = 555

Step 1: Calculate the standard error (SE)

SE=4016=404=10SE = \frac{40}{\sqrt{16}} = \frac{40}{4} = 10

Step 2: Calculate the Z-score

Z=55557510=2010=2Z = \frac{555 - 575}{10} = \frac{-20}{10} = -2

Step 3: Find the probability corresponding to the Z-score

Using the Z-table, a Z-score of -2 corresponds to a probability of 0.0228.

So, the probability that the sample mean is less than 555 is: P(xˉ<555)=0.0228P(\bar{x} < 555) = 0.0228

Final Answer for Part (a):

P(xˉ<555)=0.0228(rounded to four decimal places)P(\bar{x} < 555) = 0.0228 \quad \text{(rounded to four decimal places)}


Part (b): Sample size = 25, Find P(xˉ\bar{x} ≥ 593)

Given:

  • Population mean (μ\mu) = 575
  • Population standard deviation (σ\sigma) = 40
  • Sample size (nn) = 25
  • Desired sample mean (xˉ\bar{x}) = 593

Step 1: Calculate the standard error (SE)

SE=4025=405=8SE = \frac{40}{\sqrt{25}} = \frac{40}{5} = 8

Step 2: Calculate the Z-score

Z=5935758=188=2.25Z = \frac{593 - 575}{8} = \frac{18}{8} = 2.25

Step 3: Find the probability corresponding to the Z-score

Using the Z-table, a Z-score of 2.25 corresponds to a cumulative probability of 0.9878.

To find the probability that the sample mean is greater than or equal to 593: P(xˉ593)=10.9878=0.0122P(\bar{x} \geq 593) = 1 - 0.9878 = 0.0122

Final Answer for Part (b):

P(xˉ593)=0.0122P(\bar{x} \geq 593) = 0.0122


Summary of Answers:

  • Part (a): P(xˉ<555)=0.0228P(\bar{x} < 555) = 0.0228
  • Part (b): P(xˉ593)=0.0122P(\bar{x} \geq 593) = 0.0122

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Math Problem Analysis

Mathematical Concepts

Statistics
Normal Distribution
Sampling Distribution
Z-score
Probability

Formulas

Standard error of the mean (SE) = σ / √n
Z-score = (x̄ - μ) / SE
Probability from Z-table

Theorems

Central Limit Theorem
Properties of the Standard Normal Distribution

Suitable Grade Level

College-Level Statistics or Advanced High School (Grades 11-12)