To solve this problem using the normal approximation to the binomial distribution, we can first calculate the mean (μ) and standard deviation (σ) of the binomial distribution using the formulas:
[ \mu = np ]
[ \sigma = \sqrt{np(1-p)} ]
Where:
- ( n ) is the number of trials (400 in this case),
- ( p ) is the probability of success on each trial (0.5, as it’s a fair coin toss).
Using these formulas, we can calculate ( \mu ) and ( \sigma ):
[ \mu = 400 \times 0.5 = 200 ]
[ \sigma = \sqrt{400 \times 0.5 \times (1 – 0.5)} = \sqrt{100} = 10 ]
Now, we can use the normal distribution to approximate the probabilities:
a) More than 180 times:
We want to find ( P(X > 180) ). To use the normal approximation, we need to standardize the value:
[ z = \frac{x – \mu}{\sigma} = \frac{180 – 200}{10} = -2 ]
Using the standard normal distribution table or calculator, we find that the probability of ( Z > -2 ) is approximately 0.9772.
b) Less than 195 times:
We want to find ( P(X < 195) ). Again, we standardize the value:
[ z = \frac{x – \mu}{\sigma} = \frac{195 – 200}{10} = -0.5 ]
Using the standard normal distribution table or calculator, we find that the probability of ( Z < -0.5 ) is approximately 0.3085.
Therefore:
a) The probability of getting more than 180 heads is approximately ( 0.9772 ).
b) The probability of getting less than 195 heads is approximately ( 0.3085 ).