I NEED HELP WITH THIS Could someone PLEASE HELP ME

Question

Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p.

 

Use the given values of n and p to find the mean μ and standard deviation σ.

 

 

Use the range rule of thumb to find the minimum usual value μ−2σ and the maximum usual value μ+2σ.

 

n=1415, p=3/5 μ

 

 

What are they talking about????????

Answer 1

We'll use seeds as an example of a binomial distribution.

 

We're planting tomato seeds and the probability of a seed germinating is 0.8. We plant 9 seeds total. Let 'X' be equal to the number of tomato seeds that actually germinate.

 

Now, with that information given above, we can solve for the:

 

mean (µ): the average number of seeds that we could expect to germinate = n * p

 

standard deviation (σ): sqrt of n*p*(1-p)

 

Once you calculate the mean and standard deviation, you can calculate the minimum usual value and the max usual value.

 

 

Now, in your problem:

 

n = 1415

p = 3/5

 

so,

 

the mean would be: (1415)(3/5) = 849

 

the standard deviation would be:  sqrt (mean)(1-3/5) = 18.4282

 

 

Next, we find the minimum and maximum usual values according to the equations you gave:

 

minimum:  mean - 2(standard deviation)

                 = 849 - 2(18.4282)

                 = 812.1436

 

maximum: mean + 2(standard deviation)

                 = 849 + 2(18.4282)

                 = 885.8564

Answer 2

In a binomial distribution the mean is found using the formula mean = n times p. The standard deviaton is the square root of the variance. The binomial variance formula is var = n times p times (1-p).

 

Once you have the mean (μ) and stan dev. (σ) you can use the range rule of thumb.