Question

The weights of a certain brand of candies are normally distributed with a mean weight of

0.85550.8555

g and a standard deviation of

0.05170.0517

g. A sample of these candies came from a package containing

458458

​candies, and the package label stated that the net weight is

391.4391.4

g.​ (If every package has

458458

​candies, the mean weight of the candies must exceed

StartFraction 391.4 Over 458 EndFraction391.4458equals=0.85450.8545

g for the net contents to weigh at least

391.4391.4

​g.)

a. If 1 candy is randomly​ selected, find the probability that it weighs more than

0.85450.8545

g.The probability is

nothing.

​(Round to four decimal places as​ needed.)

Answer 1

Solution:

Given: The weights of a certain brand of candies are normally distributed with a mean weight of 0.8555 g and a standard deviation of 0.0517 g.

Sample size = n = 458

Part a) If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8545 g.

P( X > 0.8545 ) =.............?

Find z score for x = 0.8545

thus we get:

Look in z table for z = -0.0 and 0.02 and find corresponding area.

P( Z < -0.02) = 0.4920

thus

If exact answers are needed then use following step

P( X > 0.8545) = 1 - P( X < 0.8545)

and Excel command:

=1-NORM.DIST(x, mean, std_dev, cumulative)

=1-NORM.DIST(0.8545,0.8555,0.0517,TRUE)

=0.5077

That is: P( X > 0.8545) = 0.5077

Part b) If 458 candies are randomly​ selected, find the probability that their mean weight is at least 0.8545 g.

Find z score for

thus we get:

Look in z table for z = -0.4 and 0.01 and find corresponding area.

P( Z< -0.41 ) = 0.3409

thus

If exact answers are needed then use following step

P( 0.8545) = 1 - P( < 0.8545)

and Excel command:

=1-NORM.DIST(x, mean, std_dev/SQRT(n), cumulative)

=1-NORM.DIST(0.8545,0.8555,0.0517/SQRT(458),TRUE)

=0.6605

That is:

P( 0.8545) = 0.6605