Statistics question

Question

A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n independent trials of the experiment.

n=30,

p=0.97​,

x=28

P(28)=enter your response here

​(Do not round until the final answer. Then round to four decimal places as​ needed.)

Answer 1

There is a known formula for this calculation. The binomial PMF (probability mass function) describes the probability of x successes in n trials.

binomial PMF = (number of trials choose number of successes) * (probability of success)^(number of successes) * (1 - probability of success)^(number of trials - number of successes)

In this case:

Pr(x = 28) = (30 C 28) * (0.97)^28 * (1 - 0.97)^(30 - 28) = (30 C 28) * (0.97)^28 * (0.03)^2 = 0.1669 (rounded to 4 decimal places)

Answer 2

n=30

p=.97,q=1-.97=.03

x= 28

P(x=28)=P(28.5)-P(27.5)

use a normal distribution to approximate a binomial distribution

mean=xbar= np= 29.7

var=npq= 29.7(.03) = 8.91

s=sqr8.91= about 2.985

z score for x =(x-xbar)/s

= (28.5-29.7)/2.985 =about -.402

and (27.5-29.7)/2.985 = about -.737

use z tables to find area under the normal curve at the 2 z scores

then subtract the smaller area from the larger

.3438-.2306= .1132

or, more accurately just use the binomial distribution

P(x=28) =nCx(p^x)(q^(n-x)= 30C28(.97^28)(.03)^2 = about .1669

go with the last answer

P(x=28) = .1669= 16.69% probability of exactly 28 out of 30

or maybe you meant what's the P(x>/=28) =P(28,29 or 30)?

then P(at least 28) = P(28)+P(29)+P(30) = .1669 + 30(.97^29)(.03) + (.97^30) = about .9399

= 93.99% chance of at least 28 successes out of 30 independent trials