Math  /  Data & Statistics

Questionhttps://www-awy.aleks.com/alekscgi/X/Isl.exe/1o_u-lgNslkasNW8D8A9PVVfaYzvAnKbEZqvmttllkfkbx0fBm18LItV2MuqLjsTaljMWZa5_8UrnEXm8h4nR48leR0Eiar... Homework 6.4 Question 3 of 6 (1 point) I Question Attempt: 1 of Unlimited Salma 1\checkmark 1 2\checkmark 2 =3=3 4 5 6
Single Americans In a recent year, about 22%22 \% of Americans 16 years and older are single. What is the probability that in a random sample of 225 Americans 16 or older, more than 36 are single? Round the final answer to at least 4 decimal places and intermediate zz-value calculations to 2 decimal places. P(x>36)=P(x>36)= \square

Studdy Solution

STEP 1

What is this asking? Out of 225 Americans, if 22% are single, what's the chance that *more* than 36 of them are single? Watch out! We're dealing with a *greater than*, not a *less than* probability, so we'll need to be careful with our calculations!
Also, don't forget to round correctly.

STEP 2

1. Calculate the expected value and standard deviation.
2. Calculate the z-score.
3. Find the probability.

STEP 3

Let's **define** some variables!
Let nn be our **sample size**, which is n=225n = 225.
Let pp be the **probability** of a single American, which is p=0.22p = 0.22.

STEP 4

The **expected value** μ\mu (that's the average number of single Americans we'd *expect* to see in our sample) is given by μ=np\mu = n \cdot p.
So, μ=2250.22=49.5\mu = 225 \cdot 0.22 = 49.5.
On average, we'd expect to find about **49.5** single Americans in a group of 225.

STEP 5

Now, let's find the **standard deviation** σ\sigma, which tells us how spread out our data is likely to be.
The formula is σ=np(1p)\sigma = \sqrt{n \cdot p \cdot (1-p)}.
Plugging in our values, we get σ=2250.22(10.22)=2250.220.78=38.976.24\sigma = \sqrt{225 \cdot 0.22 \cdot (1-0.22)} = \sqrt{225 \cdot 0.22 \cdot 0.78} = \sqrt{38.97} \approx 6.24.
So our **standard deviation** is approximately **6.24**.

STEP 6

The **z-score** tells us how many standard deviations a particular value is away from the mean.
We want to know the probability of finding *more* than 36 single Americans.
The formula for the z-score is z=xμσz = \frac{x - \mu}{\sigma}, where xx is the value we're interested in (in this case, 36).

STEP 7

Plugging in our values, we get z=3649.56.24=13.56.242.16z = \frac{36 - 49.5}{6.24} = \frac{-13.5}{6.24} \approx -2.16.
So, 36 is approximately **-2.16** standard deviations away from the mean.

STEP 8

Now, we need to find the probability associated with this z-score.
We want P(x>36)P(x > 36), which is the same as P(z>2.16)P(z > -2.16).
Since z-tables usually give us P(z<Z)P(z < Z), we can use the fact that the total probability under the curve is 1.
So, P(z>2.16)=1P(z<2.16)P(z > -2.16) = 1 - P(z < -2.16).

STEP 9

Looking up the z-score of -2.16 in a z-table (or using a calculator), we find that P(z<2.16)0.0154P(z < -2.16) \approx 0.0154.

STEP 10

Therefore, P(z>2.16)=10.0154=0.9846P(z > -2.16) = 1 - 0.0154 = 0.9846.

STEP 11

The probability that more than 36 Americans in our sample are single is approximately **0.9846**.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord