Question

Find the probabiliy and interpet the resils. If convenient, use iechnobog to find the probsblity. sample is less than $59,5007 Assume o =$6.100. The probublify thst the mean salary of ine smple is less than 569,500is (Found bour decimal places as needed) interpet the resuls. Choose the corect answer below. D. Abore 1.457 of samples of 44 specisists wh hare a mean salary evs than 359,500 tha a not an umpual erer.

Answer 1

The probability that the mean salary of a sample of 44 specialists is less than $59,500 is approximately 0.0146.

Step-by-step explanation:

Using the z-score formula with the given information:

`\[ z = \frac{{\bar{x} - \mu}}{{\frac{\sigma}{{√(n)}}}} = \frac{{59500 - 59507}}{{\frac{6100}{{√(44)}}}} \]`

Calculating this results in a z-score of approximately -0.1717.

Next, referring to a standard normal distribution table or using technology like a statistical calculator or software, the probability corresponding to this z-score can be found. Using technology such as Python or R, one can employ functions like `pnorm` or `scipy.stats.norm.cdf` to find this probability directly from the z-score.

The probability, rounded to four decimal places, is approximately 0.0146, or 1.46%.

This means that in a normally distributed population with a mean of $59,507 and a standard deviation of $6,100, the probability of obtaining a sample mean salary of less than $59,500 from a sample of 44 specialists is about 1.46%. Therefore, it's within the range of expected variations and is not an unusually rare event. In practical terms, this suggests that observing a sample mean salary below $59,500 for a group of 44 specialists is not exceptionally uncommon, given the parameters of the population's distribution.