Question

Assume that a procedure yeilds a binomial with n trial and the probability of success for one trial is p. Use the given values of n and p to find the mean and standard deviation. Also, use the range rule of thumb to find the minimum usual value mean -2standard deviation and the maximum usual value mean + 2 standard deviation n=1490,p=2/5

Answer 1

The value of minimum usual value is, $\mu-2\sigma=-119.2$

The value of maximum usual value is, $\mu+2\sigma=1311.2$

Given the values of the parameters of Binomial Distribution are,

Total number of trials (n) = 1490

probability of success in one trial is (p) = 2/5

The probability of failure in on trial is given by,

$q=1-p=1-\frac{2}{5}=\frac{5-2}{5}=\frac{3}{5}$

For Binomial distribution we know that,

Mean $(\mu)=np=1490\times\frac{2}{5}=596$

and Standard Deviation $(\sigma)=\sqrt{npq}=\sqrt{1490\times\frac{2}{5}\times\frac{3}{5}}=357.6$

Now, calculating the required measurement we get,

The minimum usual value is given by,

Mean -2 Standard Deviation $=\mu-2\sigma=596-2\times357.6=-119.2$

The maximum usual value is given by,

Mean + 2 Standard Deviation $=\mu+2\sigma=596+2\times357.6=1311.2$

Hence the minimum and maximum usual values are -119.2 and 1311.2 respectively.

Learn more about Binomial Distribution here -

brainly.com/question/15246027

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