Question

The average number of sugar in a generic brand of cereal is 660mg, and the standard deviation is 35mg. Assume the variable is normally distributed.

a.) if a single cereal is selected, find the probability that the sugar content will be more than 670mg.

b.) if a sample of 10 cereals is selected, find the probability that the mean of the sample will be larger than 670mg.

c.)Why is the probability for part (a) greater than for part (b)?

Answer 1

Solution :

Given ,

mean = = 660

standard deviation = = 35

P(x > 670) = 1 - P(x<670 )

= 1 - P[(x -) / < (670-660) /35 ]

= 1 - P(z < 0.29)

Using z table

= 1 - 0.3859

probability = 0.6141

b.

n = 10

= 660

= / n = 35/ 10 = 11.07

P( >670 ) = 1 - P( <670 )

= 1 - P[( - ) / < (670-660) / 11.07]

= 1 - P(z <0.90 )

Using z table

= 1 - 0.8159

probability = 0.1841

c)when sample size increase area will get decreases

in part a sample is 1 and part be sample is 10 thats why  probability for part (a) greater than for part (b)