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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. 0.8141
b. 0.1859
c. 0.6554
d. 0.3446

1 Answer

6 votes

Answer:

d. 0.3446

Explanation:

We need to calculate the z-score for the given weekly income.

We calculate the z-score of $1000 using the formula


z=(x-\mu)/(\sigma)

From the question, the standard deviation is
\sigma=250 dollars.

The average weekly income is
\mu=1,100 dollars.

Let us substitute these values into the formula to obtain:


z=(1,000-1,100)/(250)


z=(-100)/(250)


z=-0.4

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,
Z_(-0.4)=0.34458.

We round to 4 decimal places to obtain:
Z_(-0.4)=0.3446.

Therefore the probability that a trainee earns less than $1,000 a week is
P(x\:<\:1000)=0.3446.

The correct choice is D.

The weekly incomes of trainees at a local company are normally distributed with a-example-1
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