The probability is \(0.314\) that the gestation period of a woman will exceed \(9\) months. In six human births, what is the probability that the number in which the gestation period exceeds \(9\) months is

a. exactly three?

b. exactly five?

c. at least five?

d. between three and five times, inclusive?

e. Determine the probability distribution of the random variable \(X\), the number of six human births in which the gestation period exceeds \(9\) months.

f. Identify the probability distribution of \(X\) as right skewed, symmetric or left skewed without consulting its probability distribution or drawing its probability histogram.

g. Draw a probability histogram for \(X\).

h. Use your answer from part (c) and definitions \(5.9\) and \(5.10\) on pages \(227\) and \(229\) respectively to obtain the mean and standard deviation of the random variable \(X\).

i. Use formula \(5.5\) on page \(239\) to obtain the mean and standard deviation of the random variable \(X\).

j. Interpret your answer for the mean in words.

The probability that the gestation period of a woman will exceed \(9\) months is \(0.314\). Consider the six human births.

From the given information it can conclude that the number of gestation periods exceeds \(9\) months follows a binomial distribution with \(n=6\) and \(p=0.314\).

Let the random variable \(X\) is that the number of gestation period exceed \(9\) months among six human births.

The probability of exactly three gestation period of woman will exceed \(9\) months is

\(P(X=3)=\binom{6}{3}0.314^{3}(1-0.314)^{6-3}\)

\(=0.1999\)

The probability of exactly five gestation period of woman will exceed \(9\) months is

\(P(X=5)=\binom{6}{5}0.314^{5}(1-0.314)^{6-5}\)

\(=0.0126\)

The probability of at least five gestation period of woman will exceed \(9\) months is

\(P(x\geq 5)=P(X=5)+P(X=6)\)

\(=\binom{6}{5}0.314^{5}(1-0.314)^{6-5}+\binom{6}{6}0.314^{6}(1-0.314)^{6-6}\)

\(=0.0126+0.0010\)

\(=0.0136\)

The probability of between three and five inclusively the gestation period of woman will exceed \(9\) months is

\(P(3\leq X\leq 5)=P(X=3)+P(X=4)+P(X=5)\)

\(=\binom{6}{3}0.314^{3}(1-0.314)^{6-3}+\binom{6}{4}0.315^{4}(1-0.314)^{6-4}+\binom{6}{5}0.314^{5}(1-0.314)^{6-5}\)

\(=0.1999+0.0686+0.0126\)

\(=0.2811\)

Evaluate the probabilities at \(x=0,1,2,3,4,5\)

\(P(X=0)=\binom{6}{0}0.314^{0}(1-0.314)^{6-0}\)

\(=0.1042\)

\(P(X=1)=\binom{6}{1}0.314^{1}(1-0.314)^{6-1}\)

\(=0.2862\)

\(P(X=2)=\binom{6}{2}0.314^{2}(1-0.314)^{6-2}\)

\(=0.3275\)

\(P(X=3)=\binom{6}{3}0.314^{3}(1-0.314)^{6-3}\)

\(=0.1999\)

\(P(X=4)=\binom{6}{4}0.314^{4}(1-0.314)^{6-4}\)

\(=0.0686\)

\(P(X=5)=\binom{6}{5}0.314^{5}(1-0.314)^{6-5}\)

\(=0.0126\)

\(P(X=6)=\binom{6}{6}0.314^{6}(1-0.314)^{6-6}\)

\(=0.0010\)

Combining this information in a table, it can obtain the probability distribution:

The binomial distribution is

• Symmetric if \(p=0.5\),
• Skewed to left if \(p>0.5\),
• Skewed to right if \(p<0.5\).

In this case, \(p=0.314<0.5\).

Therefore the distribution is skewed to right.

The height of the bars are equal to the probability and the bars are centered about the values of \(X\).

The mean

\(\mu =(0\times 0.1042+1\times 0.2862+2\times 0.3275+3\times 0.1999+4\times 0.0686+5\times 0.0126+6\times 0.0010)\)

\(=1.8843\)

The standard deviation

\(\sigma =\sqrt{0^{2}\times0.1042+1^{2}\times0.2862+2^{2}\times0.3275+3^{2}\times0.1999+4^{2}\times0.0686+5^{2}0.0126+6^{2}0.0010-1.8843^{2}}\)

\(=1.1372\)

The mean

\(\mu =6\times0.314\)

\(=1.884\)

The standard deviation

\(\sigma =\sqrt{6\times0.314\times(1-0.314)}\)

\(=1.1368\)

Note that the small difference with the results of part (h) are due to rounding error.

On average, the gestation period of woman will exceed \(9\) months in \(1.884\) out of \(6\) human births.