Question

Cans of pepsir are labeled to indicate that they contain 16oz. The population standard deviation is .15oz. A random sample of 49 cans is selected

Find the mean and standard deviations for this smpling distrubutiono

B. Find the probability that a sample of 49 cans will have a sample mean greater than 16.15oz

Answer 1

Solution :

Given that ,

mean = = 16

standard deviation = = 0.15

n = 49

sampling distribution of sample mean = = 16

sampling distribution of standard deviation = = / n = 0.15 / 49 = 0.0214

P( >16.15 ) = 1 - P( <16.15 )

= 1 - P[( - ) / < (16.15 -16) / 0.0214 ]

= 1 - P(z <7.14 )

Using z table

= 1 - 1

= 0

probability= 0