Question

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

0.85960.8596

g and a standard deviation of

0.05240.0524

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

440440

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

375.5375.5

g.​ (If every package has

440440

​candies, the mean weight of the candies must exceed

StartFraction 375.5 Over 440 EndFraction375.5440equals=0.85340.8534

g for the net contents to weigh at least

375.5375.5

​g.)

Answer 1

SOLUTION:

From given data,

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8596 g and a standard deviation of  0.0524 g. A sample of these candies came from a package containing 440 ​candies, and the package label stated that the net weight is 375.5 g.​ (If every package has  440 ​candies, the mean weight of the candies must exceed Start Fraction 375.5 Over 440 End Fraction 375.5 / 440 End Fraction equals 0.8534 g for the net contents to weigh at least 375.5 ​g.)

Where,

Normal distribution ,

mean = = 0.8596

standard deviation = = 0.0524

Sample size = n = 440

We convert this to standard normal using z = - /

Where,

= 375.5 / 440 =  0.8534

= = 0.8596

= / sqrt(n) = 0.0524 / sqrt(440) = 0.002498

By substituting all values we get,

z = 0.8534  - 0.8596 / 0.002498

z = -0.0062 / 0.002498

z = -2.48

P( >0.8534) = P(z > -2.48 )

P( >0.8534) = 1- P(z < -2.48 )

P( >0.8534) = 1- 0.00657

P( >0.8534) = 0.99343

Where,

P(z < -2.48 ) : in a z - table having area to the left of z,locate -2.4 in the left most column .Move across the row to the right under column 0.08 and get value 0.00657

P( x > 375.5) = 0.99343