The probability of winning a large stuffed animal in the ring-toss game at a fair is 10%. Find the probability of winning at least 5 of 50 games using normal approximation. I got 0.4052 but answers says it's 0.5932.
Please help!
prob(win) = .1
prob(not win) = .9
prob(at least 5 of 50)
= 1 - (prob(none of 50) + prob(1 of 50) + prob(2 of 50) + prob(3of50) + prob(4of50) )
= 1 - (.9^50 + 50(.1) (.9)^49 + 1225(.1)^2 (.9)^48 + 19600(.1)^3 (.9)^47 + 230300(.1)^4 (.9)^46 )
= .5688
check my arithmetic, I did it twice
mean = 5
sd = sqrt(5*.9) = 2.121
look for greater than 4.5 !!!!!
use eg
http://davidmlane.com/hyperstat/z_table.html
So you have to do 1 minus 0.4052 to get your answer?
Most tables give the area below the normal distribution curve on the left (<Z). In that case, yes, you need to subtract.
The link Damon gave you has the option of finding area to the right, (>Z). In that case, you do not have to subtract from 1.
Also .4052 is straight reading of Z=-0.24, rounded from -0.5/sqrt(4.5)=-.5/2.12132=0.2357.
Even with paper tables, you could refine the reading by linear interpolation .4090*.43+.4052*.57=.4068, which will give 1-.4068=.5932, the exact book answer.
To solve this problem, we can use the normal approximation to the binomial distribution. Let's break down the steps to find the probability of winning at least 5 out of 50 games:
Step 1: Calculate the expected value and standard deviation of the binomial distribution.
The expected value (mean) of a binomial distribution is given by np, where n is the number of trials and p is the probability of success in each trial. In this case, n = 50 games and p = 0.10 probability of winning.
Expected value (mean) = np = 50 * 0.10 = 5
The standard deviation (σ) of a binomial distribution is given by √(np(1-p)). In this case,
Standard deviation (σ) = √(50 * 0.10 * (1 - 0.10)) = √(5 * 0.90) = √4.5 ≈ 2.1213
Step 2: Convert the problem into a normal distribution.
To use the normal approximation, we assume that the binomial distribution can be approximated by a normal distribution. We need to determine the values of X (the number of games won) that corresponds to winning at least 5 games.
Since we want at least 5 games won, we need to find the probability of X ≥ 5.
Step 3: Standardize the values using the z-score.
To use the normal distribution, we need to standardize our values using z-scores. The formula for the z-score is:
z = (X - μ) / σ
Where X is the number of games won, μ is the mean, and σ is the standard deviation.
In this case, X = 5, μ = 5, and σ = 2.1213
z = (5 - 5) / 2.1213 = 0 / 2.1213 = 0
So we need to find P(Z ≥ 0) which is equivalent to finding the probability from the mean to the right tail.
Step 4: Find the probability using the z-table or calculator.
Using a standard normal distribution table or a calculator, we find that P(Z ≥ 0) ≈ 0.5.
Step 5: Calculate the final probability.
Since we want to find the probability of winning at least 5 games, we need to subtract the probability of winning less than 5 from 1.
Final probability = 1 - P(X < 5) = 1 - P(Z < 0) ≈ 1 - 0.5 ≈ 0.5
Therefore, the probability of winning at least 5 out of 50 games using the normal approximation is approximately 0.5.
It appears that there was an error in your calculation. The correct answer should be 0.5, not 0.4052 or 0.5932.