Question

Suppose certain coins have weights that are normally distributed with a mean of 5.629 g and a standard deviation of 0.056 g. A vending machine is configured to accept those coins with weights between 5.559 g and 5.699 g.

b. If 280 different coins are inserted into the vending​ machine, what is the probability that the mean falls between the limits of 5.559 g and 5.699 ​g?

The probability is approximately ? ​(Round to four decimal places as​ needed.)

Answer 1

Solution :

Given ,

mean = = 5.629

standard deviation = = 0.056

P(5.559 < x <5.699 ) = P[(5.559 -5.629)/ 0.056) < (x -   ) /    < (5.699 -5.629) / 0.056) ]

= P(-1.25 < z < 1.25)

= P(z <1.25 ) - P(z < -1.25)

Using z table,

=0.8944-0.1056

=0.7888

probability=0.7888

(b)n=280

= 5.629

=  / n= 0.056/ 280=0.00335

P(5.559<    <5.699 ) = P[(5.559 -5.629) / 0.00335< ( - ) / < (5.699 -5.629) /0.00335 )]

= P(-20.89 < Z <20.89 )

= P(Z < 20.89) - P(Z < -20.89)

Using z table

=1 -0

=1

probability=1