An unbiased dice is thrown 10 times. Let X be the number of sixes obtained. Find:
C) P(X<mean)
The correct answer is .485, but I don't know how to get there. I tried
To find the probability P(X < mean), we first need to determine the mean or average of the number of sixes obtained.
Since we are throwing an unbiased dice 10 times, the probability of getting a six in a single throw is 1/6. The number of sixes obtained in 10 throws will follow a binomial distribution.
The mean (μ) of a binomial distribution is given by μ = n * p, where n is the number of trials and p is the probability of success in a single trial.
In this case, n = 10 (number of throws) and p = 1/6 (probability of getting a six in a single throw). Therefore, μ = 10 * (1/6) = 10/6 ≈ 1.67.
Now we need to find P(X < μ), which means finding the probability of getting less than 1.67 sixes.
To calculate this probability, we can use the binomial probability formula or approximate it using a normal distribution to simplify the calculation. Let's use the latter approach.
Since n is large (n = 10), we can assume that X follows an approximately normal distribution with mean μ and variance σ^2 = n * p * (1 - p).
In this case, σ^2 = 10 * (1/6) * (1 - 1/6) = 10/36 ≈ 0.278.
The standard deviation (σ) is the square root of the variance, so σ ≈ √(0.278) ≈ 0.527.
Now, to find P(X < μ), we can standardize the random variable X using the z-score formula:
z = (X - μ) / σ.
For X = 1 (since we want to find P(X < μ)), the z-score becomes:
z = (1 - 1.67) / 0.527 ≈ -1.263.
To find the probability P(Z < -1.263), where Z is a standard normal random variable, we can use a standard normal distribution table or a calculator. This value is approximately 0.102.
Therefore, P(X < μ) ≈ 0.102.
Note: The given answer of 0.485 you mentioned does not seem to be correct based on the calculations above. Make sure to double-check the answer or provide additional context if needed.