A coin is weighted so that the probability of obtaining a head in a single toss is .4. If the coin is tossed 25 times, what is the probability of obtaining fewer than 10 head? For this problem, I found the mean and standard deviation first. I got 10 for the mean and 2.45 for the standard deviation. Then I used the formula p(x<10) ---> (z<10-10/2.45) = 0. Then I used the table at the back of my textbook and got 0.5000. But my textbooks says my answer is wrong. It says that the correct answer is .4207.
I'm confused. How is it .4207 and not .5000?
what method are you using to find the mean and SD, they can't be right.
Long way:
you want ...
C(25,0) (.4)^0 (.6)^25 + C(25,1)(.4)(.6^24) + .... + C(25,9((.4^9)(.6^16)
=
I used n times p for the mean and the square root of npq for the standard deviation
To find the probability of obtaining fewer than 10 heads when tossing a weighted coin 25 times, we can use the binomial distribution formula. The formula for the probability of getting exactly x "successes" in n trials, with a probability p of success in a single trial, is:
P(x) = (nCx) * p^x * (1-p)^(n-x) where (nCx) is the number of ways to choose x successes out of n trials.
In this case, we need to find the probability of obtaining fewer than 10 heads, which is equivalent to finding the probability of obtaining 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 heads.
To calculate this, we can use the binomial formula repeatedly for each value of x and sum up the probabilities. However, as 25 is a relatively large number and performing repetitive calculations can be tedious, we can use an alternative method called the normal approximation to the binomial distribution.
The normal approximation states that when n is large and both np and n(1-p) are greater than 10 (which is true in this case), the binomial distribution can be approximated by a normal distribution with mean μ = np and standard deviation σ = sqrt(np(1-p)).
Therefore, to find the probability of obtaining fewer than 10 heads, we can convert it into a normal distribution problem using the mean and standard deviation you have calculated: μ = 10 and σ = 2.45.
Now, we need to find the z-score (standardized score) for x = 10, where we want to calculate the cumulative probability up to that point.
z = (x - μ) / σ
= (10 - 10) / 2.45
= 0
The z-score of 0 means that we need to find the cumulative probability up to the mean, which is 0.5 or 50% (from the standard normal distribution table).
So, according to your calculations, you found a probability of 0.5, not 0 as you mentioned. This means you have made an error in your calculations or interpretation of the z-score from the table.
The correct answer should be 0.5, not 0.4207. Double-check your calculations or the values you used from the table to find the correct answer.