The weekly salaries of teachers in one state are normally distributed with a mean of $490 and a standard deviation of $45. What is the probability that a randomly selected teacher earns more than $525 a week?

There is a probability of 21.77% that selected teacher earns more than $525 a week.

Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

μ = 490, σ = 45:

For x > 525:

From the normal distribution table, P(x > 525) = P(z > 0.78) = 1 - P(z < 0.78) = 1 - 0.7823 = 21.77%

There is a probability of 21.77% that selected teacher earns more than $525 a week.

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21.77% probability that a randomly selected teacher earns more than $525 a week

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

What is the probability that a randomly selected teacher earns more than $525 a week?

This is 1 subtracted by the pvalue of Z when X = 525. So

has a pvalue of 0.7823

1 - 0.7823 = 0.2177

21.77% probability that a randomly selected teacher earns more than $525 a week

4

step-by-step explanation:

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

=

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

=

=

=

⇒ option b is correct.