Question

In: Statistics and Probability

According to the data released by the Chamber of Commerce of a certain city, the weekly...

According to the data released by the Chamber of Commerce of a certain city, the weekly wages of factory workers are normally distributed with a mean of $710 and a standard deviation of $50. Find the probability that a worker selected at random from the city makes the following weekly wage. (Round your answers to four decimal places.) (a) less than $710 = (b) more than $835 = (c) between $660 and $760=

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Expert Solution

Solution:
Given in the question
the weekly wages of factory workers are normally distributed with
Mean ()= $710
Standard deviation ()= $50
Solution(a)
We need to calculate the probability that a worker selected at random from the city makes the weekly wage less than $710
P(X<710) = ?
Here we will use the standard normal table. First, we will calculate Z-score which can be calculated as
Z = (X - )/ = (710-710)/50 = 0
From Z score we found P-value
P(X<710) = 0.50
So we can say that probability that a worker selected at random from the city makes the weekly wage less than $710 is 50%
Solution(b)
We need to calculate the probability that a worker selected at random from the city makes the weekly wage more than $835
P(X>835) = ?
Z = (X - )/ = (835-710)/50 = 2.5
From Z score we found P-value
P(X>835) = 0.00621
So we can say that probability that a worker selected at random from the city makes the weekly wage More than $835 is 0.62%
Solution(c)
We need to calculate the probability that a worker selected at random from the city makes the weekly wage between $660 and $760
P($660<X<$760) = P(X<760) - P(X<660)
Z = (X - )/ = (760-710)/50 = 1
Z = (660 - 710)/50 = -1
From Z score we found P-value
P($660<X<$760) = P(X<760) - P(X<660) = 0.8413 - 0.1587 = 0.6826
So we can say that the probability that a worker selected at random from the city makes the weekly wage between $660 and $760 is 68.26%.


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