Question
A survey asked 30 tenants in an apartment building how many TV sets they had in their apartment. Let the random variable upper X denote the number of televisions per apartment. The survey results are recorded in the following table:
upper x 0 1 2 3 4 or more
Frequency 1 6 14 7 2
If one tenant is selected at random, what is the probability that they have fewer than 3 television sets in their apartment? Round your answer to the hundredths place.
(1 point)
Responses
upper P left parenthesis upper X less than 3 right parenthesis approximately equals 0.23
Image with alt text: upper P left parenthesis upper X less than 3 right parenthesis approximately equals 0.23
upper P left parenthesis upper X less than 3 right parenthesis approximately equals 0.47
Image with alt text: upper P left parenthesis upper X less than 3 right parenthesis approximately equals 0.47
upper P left parenthesis upper X less than 3 right parenthesis approximately equals 0.67
Image with alt text: upper P left parenthesis upper X less than 3 right parenthesis approximately equals 0.67
upper P left parenthesis upper X less than 3 right parenthesis equals 0.70
upper x 0 1 2 3 4 or more
Frequency 1 6 14 7 2
If one tenant is selected at random, what is the probability that they have fewer than 3 television sets in their apartment? Round your answer to the hundredths place.
(1 point)
Responses
upper P left parenthesis upper X less than 3 right parenthesis approximately equals 0.23
Image with alt text: upper P left parenthesis upper X less than 3 right parenthesis approximately equals 0.23
upper P left parenthesis upper X less than 3 right parenthesis approximately equals 0.47
Image with alt text: upper P left parenthesis upper X less than 3 right parenthesis approximately equals 0.47
upper P left parenthesis upper X less than 3 right parenthesis approximately equals 0.67
Image with alt text: upper P left parenthesis upper X less than 3 right parenthesis approximately equals 0.67
upper P left parenthesis upper X less than 3 right parenthesis equals 0.70
Answers
GPT-4o mini
To find the probability that a randomly selected tenant has fewer than 3 television sets in their apartment, we need to determine the frequency of tenants with 0, 1, or 2 televisions and divide that by the total number of tenants.
From the table provided:
- Frequency of \( X = 0 \): 1
- Frequency of \( X = 1 \): 6
- Frequency of \( X = 2 \): 14
Now, let's calculate the total frequency for \( X < 3 \):
\[
\text{Total frequency for } X < 3 = 1 + 6 + 14 = 21
\]
The total number of tenants surveyed is given as 30. Therefore, we can find the probability \( P(X < 3) \):
\[
P(X < 3) = \frac{\text{Total frequency for } X < 3}{\text{Total number of tenants}} = \frac{21}{30}
\]
Now, let's simplify this fraction:
\[
P(X < 3) = \frac{21}{30} = 0.7
\]
Rounding to the hundredths place, we have:
\[
P(X < 3) = 0.70
\]
Thus, the correct answer is:
**\( P(X < 3) = 0.70 \)**.
From the table provided:
- Frequency of \( X = 0 \): 1
- Frequency of \( X = 1 \): 6
- Frequency of \( X = 2 \): 14
Now, let's calculate the total frequency for \( X < 3 \):
\[
\text{Total frequency for } X < 3 = 1 + 6 + 14 = 21
\]
The total number of tenants surveyed is given as 30. Therefore, we can find the probability \( P(X < 3) \):
\[
P(X < 3) = \frac{\text{Total frequency for } X < 3}{\text{Total number of tenants}} = \frac{21}{30}
\]
Now, let's simplify this fraction:
\[
P(X < 3) = \frac{21}{30} = 0.7
\]
Rounding to the hundredths place, we have:
\[
P(X < 3) = 0.70
\]
Thus, the correct answer is:
**\( P(X < 3) = 0.70 \)**.