Question

(e) compute ( e(x) ).
( e(x) = )
(f) compute ( \text{var}(x) ) and ( sigma ). (round your answer for ( sigma ) to two decimal places.)
( \text{var}(x) = )
( sigma = )
submit answer

1. - / 7 points

details my notes ask your teacher practice another
you may need to use the appropriate appendix table or technology to answer this question.
you are encouraged to work this problem using a binomial table.
consider a binomial experiment with ( n = 20 ) and ( p = 0.70 ). (round your answers to four decimal places.)
(a) compute ( p(11) ).
( p(11) = )
(b) compute ( p(16) ).
( p(16) = )
(c) compute ( p(x geq 16) ).
( p(x geq 16) = )
(d) compute ( p(x leq 15) ).
( p(x leq 15) = )
(e) compute ( e(x) ).
( e(x) = )
(f) compute ( \text{var}(x) ) and ( sigma ).
( \text{var}(x) = )
( sigma = )
submit answer

1. - / 7 points

details my notes ask your teacher practice another

Explanation:

Part (a): Compute \( p(11) \)

Step 1: Recall binomial probability formula

The binomial probability formula is \( p(x) = \binom{n}{x} p^x (1 - p)^{n - x} \), where \( \binom{n}{x} = \frac{n!}{x!(n - x)!} \), \( n = 20 \), \( p = 0.70 \), and \( x = 11 \).

Step 2: Calculate combination \( \binom{20}{11} \)

\( \binom{20}{11} = \frac{20!}{11!(20 - 11)!} = \frac{20!}{11!9!} = 167960 \)

Step 3: Substitute into probability formula

\( p(11) = 167960 \times (0.70)^{11} \times (0.30)^{9} \)
First, calculate \( (0.70)^{11} \approx 0.0197732674 \) and \( (0.30)^{9} \approx 0.000019683 \)
Then, \( p(11) \approx 167960 \times 0.0197732674 \times 0.000019683 \approx 0.0654 \)

Step 1: Use binomial probability formula

\( p(x) = \binom{n}{x} p^x (1 - p)^{n - x} \), \( n = 20 \), \( p = 0.70 \), \( x = 16 \)

Step 2: Calculate combination \( \binom{20}{16} \)

\( \binom{20}{16} = \binom{20}{4} = \frac{20!}{4!16!} = 4845 \)

Step 3: Substitute into formula

\( p(16) = 4845 \times (0.70)^{16} \times (0.30)^{4} \)
Calculate \( (0.70)^{16} \approx 0.003323293 \) and \( (0.30)^{4} = 0.0081 \)
Then, \( p(16) \approx 4845 \times 0.003323293 \times 0.0081 \approx 0.1304 \)

Step 1: Recall \( P(x \geq 16) = p(16) + p(17) + p(18) + p(19) + p(20) \)

Step 2: Calculate each probability

• \( p(17) = \binom{20}{17} (0.70)^{17} (0.30)^{3} \)

\( \binom{20}{17} = \binom{20}{3} = 1140 \)
\( (0.70)^{17} \approx 0.002326305 \), \( (0.30)^{3} = 0.027 \)
\( p(17) \approx 1140 \times 0.002326305 \times 0.027 \approx 0.0716 \)

• \( p(18) = \binom{20}{18} (0.70)^{18} (0.30)^{2} \)

\( \binom{20}{18} = \binom{20}{2} = 190 \)
\( (0.70)^{18} \approx 0.001628414 \), \( (0.30)^{2} = 0.09 \)
\( p(18) \approx 190 \times 0.001628414 \times 0.09 \approx 0.0277 \)

• \( p(19) = \binom{20}{19} (0.70)^{19} (0.30)^{1} \)

\( \binom{20}{19} = 20 \)
\( (0.70)^{19} \approx 0.00113989 \), \( (0.30)^{1} = 0.3 \)
\( p(19) \approx 20 \times 0.00113989 \times 0.3 \approx 0.0068 \)

• \( p(20) = \binom{20}{20} (0.70)^{20} (0.30)^{0} \)

\( \binom{20}{20} = 1 \), \( (0.70)^{20} \approx 0.000797923 \), \( (0.30)^{0} = 1 \)
\( p(20) \approx 1 \times 0.000797923 \times 1 \approx 0.0008 \)

Step 3: Sum the probabilities

\( P(x \geq 16) = 0.1304 + 0.0716 + 0.0277 + 0.0068 + 0.0008 \approx 0.2373 \)

Answer:

\( p(11) \approx 0.0654 \)

Part (b): Compute \( p(16) \)