Homework SourseThe manufacturer of Happy Time soda machines targets 065 oun
| The manufacturer of Happy Time soda machines targets 0.65 ounces of soda syrup in each bottle by its machines. The lower and upper specification limits for the ounces of soda syrup in each bottle is 0.645 ounces (lower) and 0.655 ounces (upper). Past experience has indicated that the mean amount of in each bottle is 0.649 ounces and the standard deviation is .023. If you select a random sample of 45 soda bottles, what is the probablity that the sample mean is: | |||||||||||||||
| a. | between the target and the mean of 0.649? | ||||||||||||||
| b. | between the lower specification limit and the target? | ||||||||||||||
| c. | greater than the upper specification limit? | ||||||||||||||
| d. | less than the lower specification limit? | ||||||||||||||
| e. | the probability is 90% that the sample mean will be less than what value? | ||||||||||||||
| f. | the probability is 93% that the sample mean will be greater than what value? | ||||||||||||||
| g. | The probability is 90% that the sample mean amount will be between which two values (symmetrically distributed around the mean)? | ||||||||||||||
| The manufacturer of Happy Time soda machines targets 0.65 ounces of soda syrup in each bottle by its machines. The lower and upper specification limits for the ounces of soda syrup in each bottle is 0.645 ounces (lower) and 0.655 ounces (upper). Past experience has indicated that the mean amount of in each bottle is 0.649 ounces and the standard deviation is .023. If you select a random sample of 45 soda bottles, what is the probablity that the sample mean is: | |||||||||||||||
| a. | between the target and the mean of 0.649? | ||||||||||||||
| b. | between the lower specification limit and the target? | ||||||||||||||
| c. | greater than the upper specification limit? | ||||||||||||||
| d. | less than the lower specification limit? | ||||||||||||||
| e. | the probability is 90% that the sample mean will be less than what value? | ||||||||||||||
| f. | the probability is 93% that the sample mean will be greater than what value? | ||||||||||||||
| g. | The probability is 90% that the sample mean amount will be between which two values (symmetrically distributed around the mean)? | ||||||||||||||
Solution
Mean ( u ) =0.649
Standard Deviation ( sd )=0.0343
Number ( n ) = 45
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)
a)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 0.649) = (0.649-0.649)/0.23/ Sqrt ( 45 )
= 0/0.0343
= 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(X < 0.65) = (0.65-0.649)/0.23/ Sqrt ( 45 )
= 0.001/0.0343 = 0.0292
= P ( Z <0.0292) From Standard Normal Table
= 0.51163
P(0.649 < X < 0.65) = 0.51163-0.5 = 0.0116
b)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 0.645) = (0.645-0.649)/0.23/ Sqrt ( 45 )
= -0.004/0.0343
= -0.1167
= P ( Z <-0.1167) From Standard Normal Table
= 0.45356
P(0.645 < X < 0.65) = 0.51163-0.45356 = 0.0581
c)
P(X > 0.655) = (0.655-0.649)/0.23/ Sqrt ( 45 )
= 0.006/0.034= 0.175
= P ( Z >0.175) From Standard Normal Table
= 0.4305
d)
P(X < 0.645) = (0.645-0.649)/0.23/ Sqrt ( 45 )
= -0.004/0.0343= -0.1167
= P ( Z <-0.1167) From Standard NOrmal Table
= 0.4536
e)
P ( Z < x ) = 0.9
Value of z to the cumulative probability of 0.9 from normal table is 1.282
P( x-u/s.d < x - 0.649/0.0343 ) = 0.9
That is, ( x - 0.649/0.0343 ) = 1.28
--> x = 1.28 * 0.0343 + 0.649 = 0.693
f)
P ( Z > x ) = 0.93
Value of z to the cumulative probability of 0.93 from normal table is -1.48
P( x-u/ (s.d) > x - 0.649/0.0343) = 0.93
That is, ( x - 0.649/0.0343) = -1.48
--> x = -1.48 * 0.0343+0.649 = 0.5984
