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:
a) Fewer than 10 heads?
b) Between 10 and 12 heads, inclusive?
c) More than 15 heads?
This is what I did:
I found the mean : 25 x .4= 10
Then the standard deviation: the square root of 25x.4x.6=2.45
for part a : 10-10/2.45= 0 =0.5000
for b: p(x<10) = 10-10/2.45=0=0.5000
p(x<12) = 12-10/2.45=0.82=0.2939
for c: p(x>15) = 15-10/2.45=2.04--->-2.04=0.0207
but my textbook's answer key says I'm wrong.
Your calculations for the mean and standard deviation seem correct. However, it appears that you made some mistakes in calculating the probabilities.
a) To find the probability of obtaining fewer than 10 heads, you should use the cumulative distribution function for a binomial distribution. You can use the binomcdf() function on a calculator or use the formula:
P(X < 10) = P(X = 0) + P(X = 1) + ... + P(X = 9)
Using the binomial probability formula: P(X = k) = nCk * p^k * (1-p)^(n-k)
where n is the number of trials (25), p is the probability of success (0.4), and nCk is the number of combinations of n items taken k at a time:
P(X < 10) = P(X = 0) + P(X = 1) + ... + P(X = 9)
= nC0 * p^0 * (1-p)^(n-0) + nC1 * p^1 * (1-p)^(n-1) + ... + nC9 * p^9 * (1-p)^(n-9)
Using this formula, the correct calculation for part a is:
P(X < 10) = binomcdf(25, 0.4, 9) ≈ 0.6294
b) To find the probability of obtaining between 10 and 12 heads (inclusive), you can subtract the cumulative probabilities for 9 heads or fewer from the cumulative probabilities for 12 heads or fewer:
P(10 ≤ X ≤ 12) = P(X ≤ 12) - P(X ≤ 9)
= binomcdf(25, 0.4, 12) - binomcdf(25, 0.4, 9)
≈ 0.4127
c) To find the probability of obtaining more than 15 heads, you can subtract the cumulative probability for 15 heads or fewer from 1 (since the total probability must sum up to 1):
P(X > 15) = 1 - P(X ≤ 15)
= 1 - binomcdf(25, 0.4, 15)
≈ 0.7045
I hope this helps!