Using the average and standard devation for length variable from problem 3,find the probability that the length of 35 movies is less than 120 minutes. Assume the distribution is normally distributed. interpret your result.
To find the probability that the length of 35 movies is less than 120 minutes, we can use the concept of the standard normal distribution. The given information states that the length variable follows a normal distribution, so we can use the mean and standard deviation to calculate the probability.
1. Determine the mean and standard deviation:
We are given the average (mean) and standard deviation for the length variable from problem 3. Let's assume they are μ (mean) and σ (standard deviation).
2. Convert the given values to z-scores:
A z-score represents the number of standard deviations away from the mean a particular value is. Here, we need to find the z-score for the length of 120 minutes using the formula:
z = (x - μ) / σ,
where x is the value (120 minutes), μ is the mean, and σ is the standard deviation.
3. Look up the z-score in the standard normal distribution table:
The standard normal distribution table provides the cumulative probability for different z-scores. Find the z-score from step 2 in the table to get the corresponding probability.
4. Interpret the result:
The probability you obtain represents the likelihood that the length of 35 movies randomly selected from this distribution is less than 120 minutes. It gives an indication of how common or rare such a result is within the given distribution.
Remember, to perform these calculations, you will need the values of the average and standard deviation for the length variable.