Question 1174398: The mean salary offered to students who are graduating from Coastal State University this year is $24,260, with a standard deviation of $3712. A random sample of 75 Coastal State students graduating this year has been selected. What is the probability that the mean salary offer for these 75 students is $24,000 or less?
Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **1. Calculate the Standard Error:**
* The standard error of the mean (SEM) is the standard deviation of the sampling distribution of the mean.
* SEM = σ / √n
where:
* σ = population standard deviation ($3712)
* n = sample size (75)
* SEM = 3712 / √75 ≈ 428.3193
**2. Calculate the z-score:**
* The z-score measures how many standard errors the sample mean is away from the population mean.
* z = (x̄ - μ) / SEM
where:
* x̄ = sample mean ($24,000)
* μ = population mean ($24,260)
* SEM = standard error of the mean (428.3193)
* z = (24000 - 24260) / 428.3193 ≈ -0.6070
**3. Find the Probability:**
* We want to find the probability that the sample mean is $24,000 or less, which is equivalent to finding the area to the left of the z-score of -0.6070 in the standard normal distribution.
* Using a z-table or calculator, we find that the probability is approximately 0.272.
**Therefore, the probability that the mean salary offer for these 75 students is $24,000 or less is approximately 0.272.**
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