Question

- The average hourly wage of workers at a fast food restaurant is $6.50/hr. Assume the wages are normally distributed with a standard deviation of $0.45. If a worker at this fast food restaurant is selected at random,

- What is the probability that the worker earns more than $6.75?

- What is the probability that the worker earns between $6.50 and $6.75?

Answer #1

Solution :

Given that ,

mean = = $ 6.50

standard deviation = = $ 0.45

a) P(x > 6.75 ) = 1 - P(x < 6.75 )

= 1 - P[(x - ) / < ( 6.75 - 6.50 ) / 0.45 ]

= 1 - P(z < 0.56 )

Using z table

= 1 - 0.7123

= 0.2877

b) P( 6.50 < x < 6.75 ) = P[( 6.50 - 6.50 )/ 0.45 ) < (x - ) / < ( 6.75 - 6.50 ) / 0.45 ) ]

= P( 0.00 < z < 0.56 )

= P(z < 0.56 ) - P(z < 0.00 )

Using z table

= 0.7123 - 0.5

= 0.2123

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