An automatic weighing machine used to dispense and package powered drugs can be calibrated to accurately dispense any weight between 900 and 1100 grams. It is known that the actual weights dispensed by the machine are normally distributed about the calibrated value(ie the mean value)with a standard deviation of 1.5 grams. If the machine is calibrated to 1000 grams,what percentage of the packages will weigh between 996 and 1003 grams?:eek:

Use your z score formula.

z=\frac{x-{\mu}}{\sigma}

In this case, {\mu}=1000, \;\ {\sigma}=1.5

\frac{1003-1000}{1.5}=

\frac{996-1000}{1.5}=

Find the z score for each, look up their values in the body of the table.

Then, subtract the two to find the area in between them.

would the answer be -9x10-3 in calculator?thank you for answering.:)

No, sorry. You can't have negative area

I gave you the formulas to find the z scores. Look up their corresponding values in the body of the table and subtract them.

Thank u:)

Observation of the blood pressure of a very large number of people at a certain health centre indicates that bood pressure is normally distributed.90% of the people monitored have blood pressure of less than 124.5mmHg and only 5% of the people monitored have blood pressure of less than 101mmHg.Determine the mean blood pressure,u,and the standard deviation,o.

in a survey of an assembly line manufacturing heart monitors it is found that in any 15 minute period during the morning shift the average number of faulty heart monitors produced is 0.2,and that in any 30 minute period in the afternoon shift the average number of faulty heart monitors produced is 0.25.The morning shift lasts for 3 hours and the afternoon shifts last for 4 hours.
on any particular day,what is the probability that.
a/there will be less than 3 faulty heart monitors produced during the morning shift?
b/there will be exactly 4 faulty heart monitors produced during the afternoon shift?
c/there will be only 2 faulty heart monitors produced during that day?

a/write down the binomial expansion of(p plus q)3.

a/write down the binomial expansion of(p plus q)3.

Do you mean (p+q)^{3}

Just type (p+q)^3. That 3 on the end just looks like you're multiplying by 3.

Anyway, these things follow a nice pattern.

The first one is always p^3.

The second: the exponent goes in front as a coefficient, decrease p by 1 and increase q by 1.

3p^{2}q

Now, it repeats the coefficient, p comes down 1 and q increases 1.

3pq^{2}

The last one is q^3.

So, we have p^{3}+3p^{2}q+3pq^{2}+q^{3}

If this were longer, say (p+q)^6.

To find successive coefficients, multiply the coefficient by the exponent of p and divide by 1 more than the exponent of q.

So, we have p^{6}+6p^{5}q

To find the next coefficient, we have \frac{6\cdot 5}{2}=15

p^{6}+6p^{5}q+15p^{4}q^{2}

\frac{15\cdot 4}{3}=20

p^{6}+6p^{5}q+15p^{4}q^{2}+20p^{3}q^{3}

Now, since the exponents in p and q are the same(since we had an even exponent in the beginning, 6), we begin to repeat coefficients beginning with 15 and go backwards.

Look at Pascal's triangle.