Question

use the probability distribution to complete parts (a) and (b) below.
the number of school - related extracurricular activities per student
activities 0 1 2 3 4 5 6 7
probability 0.058 0.124 0.163 0.178 0.214 0.128 0.083 0.052
(a) find the mean, variance, and standard deviation of the probability distribution.
the mean is
(round to one decimal place as needed.)

Explanation:

Step1: Recall the formula for the mean of a discrete probability distribution

The mean (expected value) \( \mu \) of a discrete probability distribution is given by \( \mu = \sum_{i} x_i P(x_i) \), where \( x_i \) are the values of the random variable and \( P(x_i) \) are their corresponding probabilities.

Step2: Identify the values of \( x_i \) and \( P(x_i) \)

Here, \( x_i \) are 0, 1, 2, 3, 4, 5, 6, 7 and their corresponding probabilities \( P(x_i) \) are 0.058, 0.124, 0.163, 0.178, 0.214, 0.128, 0.083, 0.052 respectively.

Step3: Calculate the product \( x_i P(x_i) \) for each \( i \)

- For \( x = 0 \): \( 0 \times 0.058 = 0 \)
- For \( x = 1 \): \( 1 \times 0.124 = 0.124 \)
- For \( x = 2 \): \( 2 \times 0.163 = 0.326 \)
- For \( x = 3 \): \( 3 \times 0.178 = 0.534 \)
- For \( x = 4 \): \( 4 \times 0.214 = 0.856 \)
- For \( x = 5 \): \( 5 \times 0.128 = 0.64 \)
- For \( x = 6 \): \( 6 \times 0.083 = 0.498 \)
- For \( x = 7 \): \( 7 \times 0.052 = 0.364 \)

Step4: Sum up all the products to find the mean

\( \mu = 0 + 0.124 + 0.326 + 0.534 + 0.856 + 0.64 + 0.498 + 0.364 \)
First, add 0.124 and 0.326: \( 0.124 + 0.326 = 0.45 \)
Then add 0.534: \( 0.45 + 0.534 = 0.984 \)
Then add 0.856: \( 0.984 + 0.856 = 1.84 \)
Then add 0.64: \( 1.84 + 0.64 = 2.48 \)
Then add 0.498: \( 2.48 + 0.498 = 2.978 \)
Then add 0.364: \( 2.978 + 0.364 = 3.342 \)
Rounding to one decimal place, \( \mu \approx 3.3 \) (Wait, wait, let's recalculate the sum correctly:

Wait, let's add all the terms again:

0 (from x=0) + 0.124 (x=1) = 0.124

0.124 + 0.326 (x=2) = 0.45

0.45 + 0.534 (x=3) = 0.984

0.984 + 0.856 (x=4) = 1.84

1.84 + 0.64 (x=5) = 2.48

2.48 + 0.498 (x=6) = 2.978

2.978 + 0.364 (x=7) = 3.342

Wait, 3.342 rounded to one decimal place is 3.3? Wait, no, 3.342 is closer to 3.3? Wait, 3.342, the first decimal is 3, the second is 4, so when rounding to one decimal place, we look at the second decimal. 4 is less than 5? Wait, no, 3.342: the number is 3.3 (first decimal) and 4 (second). Wait, 3.342, to round to one decimal place, we check the hundredth place (4). Since 4 < 5, we keep the tenths place as is? Wait, no, 3.342: the tenths digit is 3, hundredths is 4, thousandths is 2. So rounding to one decimal place: 3.3? Wait, but let's check the calculation again. Wait, maybe I made a mistake in the sum.

Wait, let's recalculate each \( x_i P(x_i) \):

- x=0: 0*0.058 = 0
- x=1: 1*0.124 = 0.124
- x=2: 2*0.163 = 0.326 (correct)
- x=3: 3*0.178 = 0.534 (correct)
- x=4: 4*0.214 = 0.856 (correct)
- x=5: 5*0.128 = 0.64 (correct)
- x=6: 6*0.083 = 0.498 (correct)
- x=7: 7*0.052 = 0.364 (correct)

Now sum all these:

0 + 0.124 = 0.124

0.124 + 0.326 = 0.45

0.45 + 0.534 = 0.984

0.984 + 0.856 = 1.84 (because 0.984 + 0.856 = 1.84, correct)

1.84 + 0.64 = 2.48 (correct)

2.48 + 0.498 = 2.978 (correct)

2.978 + 0.364 = 3.342 (correct)

So 3.342, when rounded to one decimal place, is 3.3? Wait, no, 3.342: the first decimal is 3, the second is 4. Since 4 < 5, we round down, so 3.3? Wait, but let's check with a calculator: 3.342 rounded to one decimal place is 3.3? Wait, no, 3.342 is 3.3 when rounded to one decimal place? Wait, 3.342: the tenths place is 3, hundredths is 4. So yes, because 4 is less than 5, we keep the tenths place as 3. So the mean is approximately 3.3? Wait, but let's check again. Wait, maybe I made a mistake in the problem. Wait, the probabilities: let's check if the sum of probabilities is 1. Let's sum the probabilities: 0.058 + 0.124 + 0.163 + 0.178 + 0.214 + 0.128 + 0.083 + 0.052.

Sum: 0.058 + 0.124 = 0.182; 0.182 + 0.163 = 0.345; 0.345 + 0.178 = 0.523; 0.523 + 0.214 = 0.737; 0.737 + 0.128 = 0.865; 0.865 + 0.083 = 0.948; 0.948 + 0.052 = 1.0. So the probabilities sum to 1, so the distribution is valid.

So back to the mean: 3.342, rounded to one decimal place is 3.3? Wait, no, 3.342 is 3.3 when rounded to one decimal? Wait, 3.342: the number is 3.3 (tenths) and 0.042 (hundredths and thousandths). Wait, no, 3.342 is 3 units, 3 tenths, 4 hundredths, 2 thousandths. So when rounding to one decimal place, we look at the hundredth digit, which is 4. Since 4 < 5, we round down, so the tenths digit remains 3. So the mean is 3.3? Wait, but let's check with a calculator: 3.342 rounded to one decimal place is 3.3. Yes.

Wait, but maybe I made a mistake in the calculation. Let's recalculate the sum of \( x_i P(x_i) \):

x=0: 0

x=1: 0.124

x=2: 0.326 (total so far: 0.124 + 0.326 = 0.45)

x=3: 0.534 (total: 0.45 + 0.534 = 0.984)

x=4: 0.856 (total: 0.984 + 0.856 = 1.84)

x=5: 0.64 (total: 1.84 + 0.64 = 2.48)

x=6: 0.498 (total: 2.48 + 0.498 = 2.978)

x=7: 0.364 (total: 2.978 + 0.364 = 3.342)

Yes, that's correct. So the mean is 3.342, which rounds to 3.3 when rounded to one decimal place.

Answer:

3.3