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The weights of cans of Ocean brand tuna are supposed to have a net weight of 6 ounces. The manufacturer tells you that the net weight is actually a Normal random variable with a mean of 6.03 ounces and a standard deviation of 0.22 ounces. Suppose that you draw a random sample of 40 cans.
Part i) Using the information about the distribution of the net weight given by the manufacturer, find the probability that the mean weight of the sample is less than 5.99 ounces. (Please carry answers to at least six decimal places in intermediate steps. Give your final answer to the nearest three decimal places. Use Rcmdr for this question.)
Probability (as a value to 3 decimal places between 0 and 1 inclusive) ______
Part ii) Use normal approximation to find the probability that more than 56% of the sampled cans are overweight, i.e. the net weight exceeds 6 ounces. (Please carry answers to at least six decimal places in intermediate steps. Give your final answer to the nearest three decimal places. Use Rcmdr for this question.)
Probability (as a value to 3 decimal places between 0 and 1 inclusive) _______

Respuesta :

Answer:

Following are the solution to the given points:

Step-by-step explanation:

For point (i):

[tex]P(X < 5.99) = P(Z < (\frac{(5.99 - 6.03)}{\frac{0.22}{\sqrt{(40)}}})[/tex]

                     [tex]= P(Z < -1.15) \\\\= 0.125[/tex]

For point (ii):

[tex]P(X> 6) = P( Z > \frac{(6 - 6.03)}{(\frac{0.22}{\sqrt{(40)}})})[/tex]

[tex]= P(Z > -0.8624 ) \\\\ = 0.806\\\\\to n = 40 \\\\ \to p = 0.806\\\\\to 56\% = 0.56\times 40 = 22.4\\\\\to u = np = 40\times 0.806 = 32.24\\\\P(X > 22.4) = P(Z > \frac{(22.4 - 32.24)}{\sqrt{(40\times 0.806\\\times (1-0.806))}})[/tex]

                    [tex]= P(Z > -3.93 ) \\\\= 1[/tex]

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