(e) compute ( e(x) ).
( e(x) = )
(f) compute ( ext{var}(x) ) and ( sigma ). (round your answer for ( sigma ) to two decimal places.)
( ext{var}(x) = )
( sigma = )
submit answer
5. - / 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 ( ext{var}(x) ) and ( sigma ).
( ext{var}(x) = )
( sigma = )
submit answer
6. - / 7 points
details my notes ask your teacher practice another

Question

(e) compute ( e(x) ).
( e(x) = )
(f) compute ( 	ext{var}(x) ) and ( sigma ). (round your answer for ( sigma ) to two decimal places.)
( 	ext{var}(x) = )
( sigma = )
submit answer
5. - / 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 ( 	ext{var}(x) ) and ( sigma ).
( 	ext{var}(x) = )
( sigma = )
submit answer
6. - / 7 points
details my notes ask your teacher practice another

Accepted Answer

### Part (a): Compute $ p(11) $
# Explanation:
## 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 $

# Answer:
$ p(11) \approx 0.0654 $

### Part (b): Compute $ p(16) $
# Explanation:
## 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 $

# Answer:
$ p(16) \approx 0.1304 $

### Part (c): Compute $ P(x \geq 16) $
# Explanation:
## 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(x \geq 16) \approx 0.2373 $

### Part (d): Compute $ P(x \leq 15) $
# Explanation:
## Step 1: Use complement rule $ P(x \leq 15) = 1 - P(x \geq 16) $

## Step 2: Substitute $ P(x \geq 16) $ from part (c)
$ P(x \leq 15) = 1 - 0.2373 = 0.7627 $

# Answer:
$ P(x \leq 15) \approx 0.7627 $

### Part (e): Compute $ E(x) $
# Explanation:
## Step 1: Recall expected value formula for binomial distribution
The expected value of a binomial distribution is $ E(x) = np $, where $ n = 20 $ and $ p = 0.70 $.

## Step 2: Calculate $ E(x) $
$ E(x) = 20 \times 0.70 = 14 $

# Answer:
$ E(x) = 14 $

### Part (f): Compute $ \text{Var}(x) $ and $ \sigma $
# Explanation:
## Step 1: Recall variance formula for binomial distribution
The variance of a binomial distribution is $ \text{Var}(x) = np(1 - p) $, and the standard deviation $ \sigma = \sqrt{\text{Var}(x)} $.

## Step 2: Calculate variance
$ \text{Var}(x) = 20 \times 0.70 \times (1 - 0.70) = 20 \times 0.70 \times 0.30 = 4.2 $

## Step 3: Calculate standard deviation
$ \sigma = \sqrt{4.2} \approx 2.05 $ (rounded to two decimal places)

# Answer:
$ \text{Var}(x) = 4.2 $
$ \sigma \approx 2.05 $

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QUESTION IMAGE

Question

Explanation:

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

Step 1: Recall binomial probability formula

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

Step 3: Substitute into probability formula

Step 1: Use binomial probability formula

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

Step 3: Substitute into formula

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

Step 2: Calculate each probability

Step 3: Sum the probabilities

Answer:

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