Math  /  Data & Statistics

QuestionUntited document -... Traditional | Stats | N... Defense / Stats | NBA.. Utah vs LA Stats \& P... Ivica Zubac /// Stats /... Home - Northern Ess... Time Remaining: 52:24 Jona Midterm Exam: Ch 1(1,2)2(1,2,3)3(1,2,3)4(1,2)5(1,2,3)6(1,2)1(1,2) \mathbf{2 ( 1 , 2 , 3 )} 3(1,2,3) 4(1,2) 5(1,2,3) 6(1,2) Question 29 of 30 (1 point) I Question Attempt: 1 of 1 19 20 21\equiv 21 22\equiv 22 23 24 25 26 27\equiv 27 28 29
Coronary bypass surgery: A healthcare research agency reported that 41%41 \% of people who had coronary bypass surgery in 2008 were over the age of 65. Thirteen coronary bypass patients are sampled.
Part 1 of 2 (a) What is the mean number of people over the age of 65 in a sample of 13 coronary bypass patients? Round the answer to two decimal places.
The mean number of people over the age of 65 is \square .
Part 2 of 2 (b) What is the standard deviation of the number of people over the age of 65 in a sample of 13 coronary bypass patients? Round the answer to four decimal places.
The standard deviation of the number of people over the age of 65 is \square .

Studdy Solution

STEP 1

What is this asking? If 41% of heart surgery patients are over 65, how many people over 65 do we expect to see in a group of 13 patients, and how much could this number typically vary? Watch out! Don't mix up the formulas for mean and standard deviation of a binomial distribution!

STEP 2

1. Calculate the mean.
2. Calculate the standard deviation.

STEP 3

We're dealing with a **binomial distribution** here!
We have a fixed number of trials (n=13n = \mathbf{13} patients) and the probability of "success" (being over 65) is the same for each patient (p=0.41p = \mathbf{0.41}).
The mean of a binomial distribution is simply npn \cdot p.
This makes intuitive sense: it's the number of trials times the probability of success on each trial.

STEP 4

Let's **calculate the mean**: μ=np=130.41=5.33\mu = n \cdot p = \mathbf{13} \cdot \mathbf{0.41} = \mathbf{5.33} So, on average, we'd expect to see about **5.33** people over 65 in a group of 13 heart surgery patients.

STEP 5

The **standard deviation** of a binomial distribution is given by the formula σ=np(1p)\sigma = \sqrt{n \cdot p \cdot (1 - p)}.
This formula tells us how spread out the data is likely to be.
The (1p)(1-p) term represents the probability of *not* being over 65.

STEP 6

Let's **plug in our values**: σ=np(1p)=130.41(10.41)=130.410.59\sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{\mathbf{13} \cdot \mathbf{0.41} \cdot (1 - \mathbf{0.41})} = \sqrt{\mathbf{13} \cdot \mathbf{0.41} \cdot \mathbf{0.59}}

STEP 7

**Continue the calculation**: σ=3.13571.7708\sigma = \sqrt{\mathbf{3.1357}} \approx \mathbf{1.7708} Rounded to four decimal places, the standard deviation is **1.7708**.
This tells us how much the actual number of patients over 65 is likely to vary around the mean of **5.33**.

STEP 8

(a) The mean number of people over the age of 65 is **5.33**.
(b) The standard deviation of the number of people over the age of 65 is **1.7708**.

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