Answer :
(a) The probability that a single jar contains more than the stated contents is approximately 0.1587.
(b) The probability that at least one jar among ten randomly selected jars contains more than the stated contents is approximately 0.8312.
(c) To ensure that 95% of all jars contain more than the stated contents with a mean of 139.3, the standard deviation would need to be changed to approximately 1.79.
(a) To find the probability that a single jar contains more than the stated contents, we can use the standard normal distribution. First, we calculate the z-score:
[tex]\[ z = \frac{137 - 139.3}{3.02} \approx -0.76159 \][/tex]
Using a standard normal distribution table or calculator, we find that the probability corresponding to this z-score is approximately 0.7787. Since we are interested in the probability that a jar contains more than the stated contents, we subtract this value from 1:
[tex]\[ P(X > 137) = 1 - 0.7787 \approx 0.2213 \][/tex]
(b) To find the probability that at least one jar among ten randomly selected jars contains more than the stated contents, we can use the complement rule. First, we find the probability that none of the jars contain more than the stated contents:
[tex]\[ P(\text{none contain more than stated contents}) = (1 - 0.2213)^{10} \approx 0.1688 \][/tex]
Then, the probability that at least one jar contains more than the stated contents is:
[tex]\[ P(\text{at least one contains more than stated contents[/tex][tex]}) = 1 - 0.1688 \approx 0.8312 \][/tex]
(c) To find the value of the standard deviation that would ensure that 95% of all jars contain more than the stated contents with a mean of 139.3, we can use the z-score formula for a given percentile in the standard normal distribution:
[tex]\[ z = \Phi^{-1}(0.95) \approx 1.645 \][/tex]
Then, we solve for the standard deviation:
[tex]\[ 1.645 = \frac{137 - 139.3}{\text{standard deviation}} \][/tex]
[tex]\[ \text{standard deviation} = \frac{139.3 - 137}{1.645} \approx 1.79 \][/tex]
