Question

1. the following table shows the number of minutes students claim they studied for the upcoming test at home.

h = number of study minutes at home	10	15	20	25	30
p(h)	0.1	0.2	0.2	0.4	0.1

the teacher gave all the students an additional 5 minutes of study time in class before they took their test. what is the standard deviation of the distribution of total study time, t?
a.) $\sigma_t = 5.83$ b.) $\sigma_t = 7.07$ c.) $\sigma_t = 7.90$ d.) $\sigma_t = 10.83$ e.) $\sigma_t = 33.99$

1. a random variable n can take on the values of 100, 200, 300, 400 or 500. the expected value of n is 245. which of the following is the best interpretation of the expected value of random variable n?

a.) a value of n randomly selected from the distribution will be less than 245 units of the mean.
b.) the values of n vary by about 245 units from the mean of the distribution.
c.) a value of n randomly selected from the distribution will be less than 245 units of the mean.
d.) for values of n repeatedly selected at random from the distribution, the mean of the selected values will approach 245.
e.) a randomly selected value of n must equal to be 245.

1. the local elementary school’s pto is going to purchase new playground equipment for the school. they’ve asked administration to keep track of the playground equipment that the students use most frequently during recess. for the last year, the administration would walk up to the recess playground at a randomly selected time and count the number of students on the swings and the number of students on the jungle gym. the probability distribution models below show the distribution of the random variables s and j, as defined below.

s = the number of students on the swings

s	0	1	2	3	4	5	6
p(s)	0.05	0.21	0.27	0.38	0.05	0.03	0.01

j = the number of students on the jungle gym

j	0	1	2	3	4	5	6
p(j)	0.05	0.06	0.18	0.35	0.30	0.05	0.01

assume that the number of students on the swings and students on the jungle gym are independent. what are the mean and standard deviation for the total number of students on both the swings and jungle gym?
a.) $\mu = 5$ children and $\sigma = 1.67$ children
b.) $\mu = 5$ children and $\sigma = 2.36$ children
c.) $\mu = 5.28$ children and $\sigma = 1.67$ children
d.) $\mu = 5.28$ children and $\sigma = 2.36$ children
e.) $\mu = 5.28$ children and $\sigma = 2.79$ children

Explanation:

Question 5

Step1: Find the standard deviation of \( h \)

First, calculate the mean \( \mu_h \) of \( h \):

$$ \mu_h = 10(0.1) + 15(0.2) + 20(0.2) + 25(0.4) + 30(0.1) = 1 + 3 + 4 + 10 + 3 = 21 $$

Then, calculate the variance \( \sigma_h^2 \):

$$ \sigma_h^2 = (10 - 21)^2(0.1) + (15 - 21)^2(0.2) + (20 - 21)^2(0.2) + (25 - 21)^2(0.4) + (30 - 21)^2(0.1) $$

$$ = (-11)^2(0.1) + (-6)^2(0.2) + (-1)^2(0.2) + (4)^2(0.4) + (9)^2(0.1) $$

$$ = 12.1 + 7.2 + 0.2 + 6.4 + 8.1 = 34 $$

So, \( \sigma_h = \sqrt{34} \approx 5.83 \)

Step2: Analyze the effect of adding 5 minutes

When we add a constant (5 minutes) to a random variable, the standard deviation remains the same because the spread of the data doesn't change. So, the standard deviation of \( T = h + 5 \) is the same as the standard deviation of \( h \), which is approximately 5.83.

The expected value of a random variable represents the long - run average of the variable when we take many independent samples. Option a is incorrect as it misinterprets the expected value. Option b is about standard deviation, not expected value. Option c is a repeat of a wrong statement. Option e is wrong because a single random selection doesn't have to equal the expected value. Option d correctly states that for values of \( N \) repeatedly selected at random from the distribution, the mean of the selected values will approach 245.

Step1: Calculate the mean of \( S \) (\( \mu_S \))

$$ \mu_S = 0(0.05)+1(0.21)+2(0.27)+3(0.38)+4(0.05)+5(0.03)+6(0.01) $$

$$ = 0 + 0.21+0.54 + 1.14+0.2 + 0.15+0.06 = 2.3 $$

Step2: Calculate the mean of \( J \) (\( \mu_J \))

$$ \mu_J=0(0.05)+1(0.06)+2(0.18)+3(0.35)+4(0.30)+5(0.05)+6(0.01) $$

$$ =0 + 0.06+0.36 + 1.05+1.2+0.25+0.06 = 2.98 $$

Step3: Calculate the mean of \( S + J \) (\( \mu_{S + J} \))

Since \( S \) and \( J \) are independent, \( \mu_{S+J}=\mu_S+\mu_J=2.3 + 2.98 = 5.28 \)

Step4: Calculate the variance of \( S \) (\( \sigma_S^2 \))

First, calculate \( E(S^2) \):

$$ E(S^2)=0^2(0.05)+1^2(0.21)+2^2(0.27)+3^2(0.38)+4^2(0.05)+5^2(0.03)+6^2(0.01) $$

$$ =0 + 0.21+1.08+3.42+0.8+0.75+0.36 = 6.62 $$

$$ \sigma_S^2=E(S^2)-\mu_S^2=6.62-(2.3)^2=6.62 - 5.29 = 1.33 $$

Step5: Calculate the variance of \( J \) (\( \sigma_J^2 \))

First, calculate \( E(J^2) \):

$$ E(J^2)=0^2(0.05)+1^2(0.06)+2^2(0.18)+3^2(0.35)+4^2(0.30)+5^2(0.05)+6^2(0.01) $$

$$ =0+0.06 + 0.72+3.15+4.8+1.25+0.36 = 10.34 $$

$$ \sigma_J^2=E(J^2)-\mu_J^2=10.34-(2.98)^2=10.34 - 8.8804 = 1.4596 $$

Step6: Calculate the variance of \( S + J \) (\( \sigma_{S + J}^2 \))

Since \( S \) and \( J \) are independent, \( \sigma_{S + J}^2=\sigma_S^2+\sigma_J^2=1.33 + 1.4596 = 2.7896 \)
Then, \( \sigma_{S + J}=\sqrt{2.7896}\approx1.67 \)

Answer:

a.) \( \sigma_t = 5.83 \)

Question 6