A die is rolled 180 times. You want to use normal approximation to find the probability that the number of 4 was rolled at least 100 times.

Question

1) What is the mean of the normal distribution that you would use in this case.



2) What is the standard deviation of the normal distribution that you would use in this case.

Answer 1

To approximate the number of times a specific outcome occurs when rolling a die 180 times, we can model the situation using a binomial distribution.

1. Mean of the Normal Distribution: For a binomial distribution with parameters \( n \) (the number of trials) and \( p \) (the probability of success), the mean \(\mu\) is calculated using the formula:

\[ \mu = n \cdot p \]

In this case, we have:

• \( n = 180 \) (the number of rolls)
• \( p = \frac{1}{6} \) (the probability of rolling a 4, since there is one favorable outcome out of six possible outcomes)

Calculating the mean:

\[ \mu = 180 \cdot \frac{1}{6} = 30 \]

2. Standard Deviation of the Normal Distribution: The standard deviation \(\sigma\) for a binomial distribution is given by the formula:

\[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} \]

Calculating the standard deviation:

\[ \sigma = \sqrt{180 \cdot \frac{1}{6} \cdot \left(1 - \frac{1}{6}\right)} = \sqrt{180 \cdot \frac{1}{6} \cdot \frac{5}{6}} \]

First, calculate \(180 \cdot \frac{1}{6} \cdot \frac{5}{6}\): \[ = 180 \cdot \frac{5}{36} = 25 \]

Now taking the square root: \[ \sigma = \sqrt{25} = 5 \]

Thus, the answers are:

1. The mean is 30.
2. The standard deviation is 5.