Question

A sample of 100 is drawn from a population with a proportion equal to 0.50. Determine the probability of observing between 43 and 64 successes.

Answer 1

The probability of observing between 43 and 64 successes=0.93132

Explanation:

We are given that

n=100

p=0.50

We have to find the probability of observing between 43 and 64 successes.

Let X be the random variable which represent the success of population.

It follows binomial distribution .

Therefore,

Mean,
`\mu=np=100* 0.50=50`

Standard deviation ,
`\sigma=√(np(1-p))`

`\sigma=\sqrt{100* 0.50(1-0.50)]`

`\sigma=5`

Now,

`P(43\leq x\leq 64)=P(42.5\leq x\leq 64.5)`

`P(42.5\leq x\leq 64.5)=P((42.5-50)/(5)\leq Z\leq (64.5-50)/(5))`

`=P(-1.5\leq Z\leq 2.9)`

`P(42.5\leq x\leq 64.5)=P(Z\leq 2.9)-P(Z\leq- 1.5)`

`P(42.5\leq x\leq 64.5)=0.99813-0.06681`

`P(43\leq x\leq 64)=0.93132`

Hence, the probability of observing between 43 and 64 successes=0.93132