Question

1. now form a frequency distribution for the data. add more rows if needed. (3 points)

hours of sleep (x) frequency (f)
4 2
5 3
6 7
7 6
8 8
9 3
10 1
11 0
calculate probabilities

1. find the following probabilities. (3 points)

a. p(x < 8) =
b. p(x > 8) =

1. if three individuals are selected from the sample, without replacement, what is the probability that they all sleep less than 8 hours? (3 points)
2. if three individuals are selected from the sample, without replacement, what is the probability that at least one of them sleeps less than 8 hours? (4 points)

Explanation:

Step1: Calculate total frequency

First, find the sum of all frequencies. $2 + 3+7 + 6+8+3 + 1+0=30$.

Step2: Calculate $P(x < 8)$

Sum the frequencies of $x = 4,5,6,7$. $2+3 + 7+6=18$. Then $P(x < 8)=\frac{18}{30}=\frac{3}{5}$.

Step3: Calculate $P(x>8)$

Sum the frequencies of $x = 9,10,11$. $3 + 1+0=4$. Then $P(x>8)=\frac{4}{30}=\frac{2}{15}$.

Step4: Calculate probability for question 13

The number of individuals who sleep less than 8 hours is 18. The total number of individuals is 30.
The probability that the first - selected individual sleeps less than 8 hours is $\frac{18}{30}$.
For the second - selected individual (without replacement), the probability is $\frac{17}{29}$.
For the third - selected individual (without replacement), the probability is $\frac{16}{28}$.
The probability that all three sleep less than 8 hours is $\frac{18}{30}\times\frac{17}{29}\times\frac{16}{28}=\frac{2448}{24360}=\frac{204}{2030}\approx0.101$.

Step5: Calculate probability for question 14

The probability that none of them sleeps less than 8 hours is the probability that all three sleep 8 hours or more. The number of individuals who sleep 8 hours or more is $8 + 3+1+0 = 12$.
The probability that the first - selected individual sleeps 8 hours or more is $\frac{12}{30}$.
For the second - selected individual (without replacement), the probability is $\frac{11}{29}$.
For the third - selected individual (without replacement), the probability is $\frac{10}{28}$.
The probability that none of them sleeps less than 8 hours is $\frac{12}{30}\times\frac{11}{29}\times\frac{10}{28}=\frac{1320}{24360}=\frac{11}{203}$.
The probability that at least one of them sleeps less than 8 hours is $1-\frac{11}{203}=\frac{203 - 11}{203}=\frac{192}{203}\approx0.946$.

Answer:

a. $\frac{3}{5}$
b. $\frac{2}{15}$

1. $\frac{204}{2030}$
2. $\frac{192}{203}$