A binomial probability distribution has p=.20\ and\ n=100. a. what ar

geduiwelh 2021-09-27 Answered
A binomial probability distribution has \(\displaystyle{p}={.20}\ {\quad\text{and}\quad}\ {n}={100}\).
a. what are the mean and standard deviation?
b. Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain.
c. what is the probability of exactly 24 successes?
d. what is the probability of 18 to 22 successes?
e. what is the probability of 15 or fewer successes?

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Expert Answer

brawnyN
Answered 2021-09-28 Author has 19022 answers

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.

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