Question

How to use TI-84 to solve, Consider the probability distribution of a random variable X shown below:
f(x) = \binom13x 0.25^x 0.75^13−x,   x = 0,1,2,…13.

What are the mean (μ) and variance (σ²) of the random variable X?
μ = 2.9, and σ² = 2.4375.
μ = 3.25, and σ² = 2.4375.
μ = 2.9, and σ² = 2.9.
μ = 3.25, and σ² = 2.9.

Find the probability:   P(X < 7)?
.

Answer 1

Given :

n =13 , p = 0.25 , q= 1- p =0 .75

Formula : _nC_x * p^x * q^n-x

In a binomial distribution

Mean() = n*p = 13*0.25 = 3.25

Variance(²) = n*p*q = 13*0.25*0.75 = 2.4375

Hence we get mean=3.25 and variance=2.4375

Now, we want to calculate p(x < 7 ) that means 0,1,2,3,4,5 and 6

So, p( x < 7) = p( x <=6)  

Now note that when we use less than equal to size for value of x we can use command binomcdf , if we have to calculate for equal to sign for value of x then we use binompdf.

So, here we use binomcdf(

Using ti-84 calculator use below steps.

1) press button "2nd" --》 "VARS" --》 B: binomcdf( and hit enter.

2) plug the values

Trials(n) = 13

P= 0.25

X = 6

Paste

And hit enter ..

Hence we get p(x < 7) = 0.9757 ( Rounding 4 decimal)