Rishi S. answered • 06/26/23
Stanford Undergrad Specializing In Math and Test Prep
This problem relates to the binomial distribution, which is used when there are exactly two mutually exclusive outcomes of a trial, often referred to as "success" and "failure". In this case, the success is winning the game, and we're told the probability of success, denoted as p, is 0.19.
a) **If you play the game 923 times, what is the most likely number of wins?**
The expected number of wins (the most likely number) can be found using the formula for the mean (μ) of a binomial distribution, which is n*p, where n is the number of trials (in this case, games played) and p is the probability of success.
μ = n*p
μ = 923 * 0.19 ≈ 175.37
Rounded to one decimal place, the most likely number of wins is 175.4.
b) **Find the standard deviation for the probability distribution of X.**
The formula for the standard deviation (σ) of a binomial distribution is sqrt(n*p*q), where q is the probability of failure (which is 1 - p).
σ = sqrt(n*p*q)
σ = sqrt(923 * 0.19 * (1 - 0.19)) ≈ 12.15
Rounded to two decimal places, the standard deviation is 12.15.
c) **The range rule of thumb specifies that the minimum usual value for a random variable is μ-2σ and the maximum usual value is μ+2σ. You already found μ and σ for the random variable X.**
The minimum and maximum usual values are calculated as follows:
Minimum usual value = μ - 2σ = 175.37 - 2*12.15 ≈ 151.07
Maximum usual value = μ + 2σ = 175.37 + 2*12.15 ≈ 199.67
d) **Use the range rule of thumb to find the usual range of X values. Enter answer as an interval using square-brackets and only whole numbers.**
The question asks for the usual range of X values as an interval using only whole numbers. In this case, we round the minimum value up and the maximum value down to get the range of whole numbers:
Minimum usual value = ceil(151.07) = 152
Maximum usual value = floor(199.67) = 199
Therefore, the usual range of X values is [152, 199].
