Question

Suppose we have a binomial experiment with n= 40 trials and a probability of success p= 0.50.Is it appropriate to use a normal approximation normal distribution? Why?Compute µ and of the approximation normal distribution.Use a continuity correction factor to convert the statement r≥23 successes to a statement about the corresponding normal variable x.Estimate P(r≥23)Is it unusual for a binomial experiment with 40 trials and probability of success to 0.50 to have a 23 or more successes? Explain.

Answer 1

Answer:

We can approximate to normal.

The parameters of the approximate normal distribution are:

$\mu=np=40*0.5=20\\\\ \sigma=\sqrt{npq} =\sqrt{40*0.5*0.5} =\sqrt{10} =3.16$

$P_B(r>23)\approx 0.314$

It is not unusual to have 23 or more successes. The probabiltity of this event happening is 13.4%.

Step-by-step explanation:

To approximate the binomial distribution to a normal distribution, we can calculate n*p and n*q and if both are bigger than 5 we can do the approximation.

$n*p=n*q=0.5*40=20$

We can approximate to normal.

The parameters of the approximate normal distribution are:

$\mu=np=40*0.5=20\\\\ \sigma=\sqrt{npq} =\sqrt{40*0.5*0.5} =\sqrt{10} =3.16$

The continuity correction factor is used because the binomial is a discrete function and the normal a continous function.

The probability of 23 successes must be expressed in the normal distribution as:

$P_B(r>23)=P_N(r>23+0.5)=P_N(r>23.5)\\\\z=(23.5-20)/3.16=1.1076\\\\P(r>23.5)=P(z>1.1076)=0.134\\\\P_B(r>23)\approx 0.314$

It is not unusual to have 23 or more successes. The probabiltity of this event happening is 13.4%.