Question

QUESTION 4

The weigh of cans of salmon is randomly distributed with mean =6.05 and standard deviation = .18.  The sample size is 36. What is the probability that the mean weight of the sample is more than 6.02?  

4 decimals

QUESTION 5

1. The weigh of cans of salmon is randomly distributed with mean =6.05 and standard deviation = .18.  The sample size is 36. What is the probability that the mean weight of the sample is between 5.99 and 6.07?  

4 decimals

QUESTION 6

1. The weigh of cans of salmon is randomly distributed with mean =8 and standard deviation = 0.673.  The sample size is 98. Find the Xbar value so that only 5% of the mean weight could be more than this Xbar value (case 4).

4 decimals

Answer 1

Solution :

mean = = 6.05

standard deviation = = 0.18

n = 36

= = 6.05  

= / n = 0.18 / 36 = 0.03

4) P( > 6.02) = 1 - P( < 6.02)

= 1 - P[( - ) / < (6.02 - 6.05) / 0.03]

= 1 - P(z < -1.00)

Using z table,    

= 1 - 0.1587

= 0.8413

5) P(5.99 < < 6.07)  

= P[(5.99 - 6.05) /0.03 < ( - ) / < (6.07 - 6.05) / 0.03)]

= P(-2.00 < Z < 0.67)

= P(Z < 0.67) - P(Z < -2.00)

Using z table,  

= 0.7486 - 0.0228   

= 0.7258

6) Given that,

mean = = 8

standard deviation = = 0.673

n = 98

=   = 8

= / n = 0.673 / 98 = 0.06798

Using standard normal table,

P(Z > z) = 5%

= 1 - P(Z < z) = 0.05

= P(Z < z ) = 1 - 0.05

= P(Z < z ) = 0.95

= P(Z < 1.645 ) = 0.95

z = 1.645

Using z-score formula  

= z * +

= 1.645 * 0.06798 + 8

= 8.1118