Question

assume that a procedure yields a binomial distribution with ( n = 28 ) trials and the probability of success for one trial is ( p = 0.42 ).
find the mean for this binomial distribution.
(round answer to one decimal place.)
( mu = )
find the standard deviation for this distribution.
(round answer to two decimal places.)
( sigma = )
will the probability histogram for the binomial random variable be approximately bell shaped?
( \bigcirc ) no, this is bell shaped since ( np ) and ( nq ) are both ( > 5 )
( \bigcirc ) yes, this is bell shaped since ( np ) and ( nq ) are both ( leq 5 )
( \bigcirc ) yes, this is bell shaped since ( np ) and ( nq ) are both ( > 5 )
( \bigcirc ) no, this is not bell shaped since ( np ) or ( nq ) are ( leq 5 )

Explanation:

Part 1: Find the mean (\(\mu\)) of the binomial distribution

Step 1: Recall the formula for the mean of a binomial distribution

The formula for the mean (\(\mu\)) of a binomial distribution is \(\mu = n \times p\), where \(n\) is the number of trials and \(p\) is the probability of success on a single trial.

Step 2: Identify the values of \(n\) and \(p\)

From the problem, we have \(n = 28\) and \(p = 0.42\).

Step 3: Calculate the mean

Substitute the values of \(n\) and \(p\) into the formula: \(\mu = 28 \times 0.42\)
\(\mu = 11.76\)

Step 4: Round to one decimal place

Rounding \(11.76\) to one decimal place gives \(11.8\).

Part 2: Find the standard deviation (\(\sigma\)) of the binomial distribution

Step 1: Recall the formula for the standard deviation of a binomial distribution

The formula for the standard deviation (\(\sigma\)) of a binomial distribution is \(\sigma = \sqrt{n \times p \times (1 - p)}\), where \(n\) is the number of trials, \(p\) is the probability of success, and \(1 - p\) is the probability of failure.

Step 2: Identify the values of \(n\), \(p\), and \(1 - p\)

We know \(n = 28\), \(p = 0.42\), so \(1 - p = 1 - 0.42 = 0.58\).

Step 3: Calculate \(n \times p \times (1 - p)\)

Substitute the values into the formula: \(28 \times 0.42 \times 0.58\)
First, calculate \(28 \times 0.42 = 11.76\)
Then, calculate \(11.76 \times 0.58 = 6.8208\)

Step 4: Take the square root

Take the square root of \(6.8208\): \(\sigma = \sqrt{6.8208}\)
\(\sigma \approx 2.61\) (rounded to two decimal places)

Part 3: Determine if the probability histogram is approximately bell - shaped

Step 1: Recall the rule for the normal approximation to the binomial distribution

A binomial distribution can be approximated by a normal (bell - shaped) distribution if \(np\geq5\) and \(n(1 - p)\geq5\).

Step 2: Calculate \(np\) and \(n(1 - p)\)

We already calculated \(np = 28\times0.42=11.76\) and \(n(1 - p)=28\times0.58 = 16.24\)

Step 3: Check the conditions

Since \(np = 11.76>5\) and \(n(1 - p)=16.24>5\), the binomial distribution can be approximated by a normal distribution, so the probability histogram will be approximately bell - shaped.

Answer:

s:

• Mean (\(\mu\)): \(11.8\)
• Standard Deviation (\(\sigma\)): \(2.61\)
• Bell - shaped? Yes, this is bell shaped since \(np\) and \(n(1 - p)\) are both \(> 5\) (the option "Yes, this is bell shaped since \(np\) and \(n(1 - p)\) are both \(> 5\)")