Step 1

a. The binomial probability distribution has \(\displaystyle{p}={0.20}\ {\quad\text{and}\quad}\ {n}={100}\).

Assume that x is a binomial random variable with \(\displaystyle{p}={0.20}\ {\quad\text{and}\quad}\ {n}={100}\).

The formula for mean in binomial distribution is,

\(\displaystyle\mu={n}{p}\),

Where p is the probability of success and n is the number of trials.

The formula for standard deviation is,

\(\displaystyle\sum=\sqrt{{{n}{p}}}{\left({1}-{p}\right)}\).

The value of mean is,

\(\displaystyle\mu={100}\times{0.20}={20}\)

Thus, the mean is 20.

The value of standard deviation is,

\(\displaystyle\sigma=\sqrt{{{100}\times{\left({0.20}\times{0.80}\right)}}}=\sqrt{{{16}}}={4}\)

Thus, the standard deviation is 4.

Step 2

b. The requirements to check whether the normal distribution can be used to approximate the binomial distribution are:

Condition 1: \(\displaystyle{n}{p}\geq{5}\)

Condition 2: \(\displaystyle{n}{\left({1}-{p}\right)}\geq{5}\)

Condition 1: \(\displaystyle{n}{p}\geq{5}\)

Substitute 100 for n and 0.20 for p in the np.

\(\displaystyle{n}{p}={100}\times{0.20}={20}{>}{5}\)

Thus, the requirement \(\displaystyle{n}{p}{\left(={20}\right)}\geq{5}\) is satisfied.

Condition 2: \(\displaystyle{n}{\left({1}-{p}\right)}\geq{5}\)

Substitute 100 for n and 0.20 for p in the \(\displaystyle{n}{\left({1}-{p}\right)}\).

\(\displaystyle{n}{\left({1}-{p}\right)}={100}\times{\left({1}-{0.20}\right)}={100}\times{0.80}={80}{>}{5}\)

Thus, the requirement \(\displaystyle{n}{q}{\left(={80}\right)}\geq{5}\) is satisfied.

Thus, the normal distribution can be used to approximate the binomial distribution.

Step 3

c. Continuity correction:

The binomial probability is converted to a normal distribution probability by using the continuity correction.

If the binomial probability represents “exactly c” then the normal probability is \(\displaystyle{P}{\left({c}-{0.5}{<}{x}{<}{c}+{0.5}\right)}\) for any number c.

By using continuity correction, the value 0.5 is added to 24 and subtracted from 24.

That is,

\(\displaystyle{P}{\left({x}={24}\right)}={P}{\left({24}-{0.5}{<}{x}{<}{24}+{0.5}\right)}={P}{\left({23.5}{<}{x}{<}{24.5}\right)}\)

Thus, the binomial probability to a normal probability by using continuity correction is \(\displaystyle{P}{\left({23.5}{<}{x}{<}{24.5}\right)}\).

Now,

\(\displaystyle{P}{\left({23.5}{<}{x}{<}{24.5}\right)}={P}{\left({x}{<}{24.5}\right)}−{P}{\left({x}{<}{23.5}\right)}\)

Probability value:

Software Procedure:

Step-by-step procedure to obtain probability value using the EXCEL software is given below:

Open an EXCEL sheet and select the cell A1.

Enter the formula =NORM.DIST(24.5,20,4,TRUE)-NORM.DIST(23.5,20,4,TRUE) in the cell A1.

Press Enter.

Hence, \(\displaystyle{P}{\left({23.5}{<}{x}{<}{24.5}\right)}={0.0605}\).

Thus, the probability of exactly 24 successes is 0.0605.