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

Answer:

The probability of observing between 43 and 64 successes=0.93132

Step-by-step 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,$\mu=np=100\times 0.50=50$

Standard deviation , $\sigma=\sqrt{np(1-p)}$

$\sigma=\sqrt{100\times 0.50(1-0.50)]$

$\sigma=5$

Now,

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

$P(42.5\leq x\leq 64.5)=P(\frac{42.5-50}{5}\leq Z\leq \frac{64.5-50}{5})$

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

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

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

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

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