Question

A confectionery company producers’ 100 g “Real Chocolate” bars that have a mean chocolate content of 70g with a standard deviation of 0.8g. The variation in chocolate content is normally distributed. What is the probability that a chocolate bar chosen at random contains:

a) More than 71g?

b) More than 68g?

c) Less than 70g?

d) Between 69 and 72g?

Answer 1

Solution :

Given that,

mean = = 70g

standard deviation = =0.8g

A ) P ( x > 71)

= 1 - P (x < 71 )

= 1 - P ( x -  / ) < ( 71 - 70 / 0.8)

= 1 - P ( z < 1 / 0.8 )

= 1 - P ( z < 1.25)

Using z table

= 1 - 0.8944

= 0.1056

Probability = 0.1056

B ) P ( x > 68 )

= 1 - P (x < 68 )

= 1 - P ( x -  / ) < ( 68 - 70 / 0.8)

= 1 - P ( z < - 2 / 0.8 )

= 1 - P ( z < 2.5)

Using z table

= 1 - 0.9938

= 0.0062

Probability = 0.0062

C ) P ( x < 70 )

P ( x -  / ) < ( 70 - 70 / 0.8)

P ( z < 0 / 0.8 )

P ( z < 0 )

Using z table

= 0.5000

Probability = 0.5000

D ) P ( 69 < x < 72 )

P ( 69 - 70 / 0.8) < ( x -  / ) < ( 72 - 70 / 0.8)

P ( - 1 / 0.8 < z < 2 / 0.8 )

P ( -1.25 < z < 2.5)

P ( z < 2.5 - P ( z < -1.25 ) )

Using z table

= 0.9938 - 0.1056

= 0.8882

Probability = 0.8882