The number of M&M's in a package normally distributed with a mean of 48 candies and...

The number of M&M's in a package normally distributed with a mean of 48 candies and a standard deviation of 3. Lets say that we get 40 packages of M&M's. What is the probability that the average number of candies in those 40 packages is between 46 and 49?

0.49999

0.9825

0.3781

0.96499

Solutions

Expert Solution

Solution :

Given that ,

mean = = 48

standard deviation = = 3

n = 40

= 48

= / $\sqrt$ n= 3/ $\sqrt$ 40 =0.474

P(46< <49 ) = P[(46-48) /0.474 < ( - ) / < (49-48) /0.474 )]

= P(-4.26 < Z <2.13 )

= P(Z < 2.11) - P(Z <-4.26 )

Using z table

=0.9825-0

probability=0.9825

orchestra answered 1 year ago

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