The weekly incomes of trainees at a local company are normally distributed with a mean of $1,100 and a standard deviation of $250. If a trainee is selected at random, what is the probability that he or she earns less than $1,000 a week?Select one:a. 0.8141b. 0.1859c. 0.6554d. 0.3446

Answer:d. 0.3446Step-by-step explanation:We need to calculate the z-score for the given weekly income.We calculate the z-score of $1000 using the formula [tex]z=\frac{x-\mu}{\sigma}[/tex]From the question, the standard deviation is [tex]\sigma=250[/tex] dollars.The average weekly income is [tex]\mu=1,100[/tex] dollars.Let us substitute these values into the formula to obtain:[tex]z=\frac{1,000-1,100}{250}[/tex][tex]z=\frac{-100}{250}[/tex][tex]z=-0.4[/tex]We now read from the standard normal distribution table the area that corresponds to a z-score of -0.4.From the standard normal distribution table, [tex]Z_{-0.4}=0.34458[/tex].We round to 4 decimal places to obtain: [tex]Z_{-0.4}=0.3446[/tex].Therefore the probability that a trainee earns less than $1,000 a week is [tex]P(x\:<\:1000)=0.3446[/tex].The correct choice is D.