The weights of bags filled by a machine are normally distributed with a standard deviation of 0.055 kilograms and a mean that can be set by the operator. At what level should the mean weight be set if it required that only 1% of the bags weigh less than 10.5 kilograms? Round the answer to 2 decimal places.

Step-by-step explanation:

Given that the  weights of bags filled by a machine are normally distributed with a standard deviation of 0.055 kilograms and a mean that can be set by the operator.

Let the mean be M.

Only 1% of the bags weigh less than 10.5 kilograms

i.e. P(X<10.5) = 0.01

corresponding Z value for P(Z<z) = 0.01 is -0.025

i.e. 10.5 = M-0.025(0.055)

Solve for M from the above equation

M =

Rounding off we get

10.50 kgs

Mean weight should be fixed as 10.50 kg.

step-by-step explanation:

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is           symmetric about the line y = 120,000.

step-by-step explanation:

given that megan owns a clothing boutique.

the profits of 2010, 11, are recorded and shown as follows:

years since     2010 0     1     2     3       4       5

profits made in 1000's   39       84     111     120   111     84

we find that profit gradually increases form 2010 to 2013 upto 120,   then starts decreasing.

the profits for 2012 and 2014 are the same while that of 2011 and 2015 are also the same.

this shows that there is symmetry about the line profit = 120000

hence answer is:

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is           symmetric about the line y = 120,000.

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