Suppose a fair coin is tossed five times and let x be the number that turned up in those five times.
**the mean and the standard deviation are respectively...
a.) 0.5, 0.5
b.) 2.5, 0.5
c.) 2.5, 2.24
d.) 2.5, 1.12
**find P(x≥1)
a.) 0.5
b.) 1.0
c.) 0.03
d.) 0.97
* i just don't get how i can do the problem not given any of the x values
To find the mean and standard deviation, we need to understand the concept of binomial distribution. In this case, a binomial distribution is used to represent the random variable x, which represents the number of times a certain outcome (heads or tails) occurs in a given number of trials (coin tosses).
The mean of a binomial distribution is given by the formula μ = n * p, where n is the number of trials and p is the probability of the outcome occurring in each trial. In this case, since we have a fair coin, the probability of getting heads (let's define it as success) or tails (failure) is 0.5. So, the mean is μ = 5 * 0.5 = 2.5.
The standard deviation of a binomial distribution is calculated using the formula σ = √(n * p * (1 - p)). Plugging in the values, we have σ = √(5 * 0.5 * (1 - 0.5)) = 0.5.
Therefore, the correct answer for the mean and standard deviation is option c.) 2.5, 2.24.
Now let's move on to finding P(x≥1), which represents the probability of getting at least one head in five coin tosses. To solve this, we can find the complement of the event "not getting any heads", which is the same as getting all tails.
The probability of getting tails in one toss is 1 - p = 1 - 0.5 = 0.5. Since we have five tosses, the probability of getting all tails is (0.5)^5 = 0.03125.
Therefore, P(x≥1) = 1 - P(no heads) = 1 - 0.03125 = 0.96875.
Hence, the correct answer for P(x≥1) is option d.) 0.97.