Commuters were asked how many times a week they stop for coffee on their way to work. A survey found a mean of 3 times a week with a standard deviation of 0.55. Find the probability that the sum of 121 values is less than 350.
To solve this problem, we need to use the Central Limit Theorem, which states that the sum (or average) of a large number of independent and identically distributed random variables will be approximately normally distributed, regardless of the shape of the original distribution.
In this case, the original distribution we are interested in is the number of times commuters stop for coffee on their way to work, which has a mean of 3 times a week and a standard deviation of 0.55.
First, we need to find the mean and standard deviation of the sum of 121 values. Since the mean of each value is 3, the mean of the sum of 121 values would be 121 times the mean of each value, which is 121 * 3 = 363.
Next, we need to calculate the standard deviation of the sum of 121 values. Since the values are independent and identically distributed, the standard deviation of the sum is the square root of the product of the number of values and the variance of each value. The variance of each value is the square of the standard deviation, so the variance of each value is 0.55 * 0.55 = 0.3025. Therefore, the standard deviation of the sum is the square root of (121 * 0.3025) = 6.605.
Now, we can use the normal distribution to find the probability that the sum of 121 values is less than 350. We need to standardize the value 350 using the formula z = (x - mean) / standard deviation, where x is the value we want to standardize, the mean is 363, and the standard deviation is 6.605. Plugging in these values, we get z = (350 - 363) / 6.605 = -1.968.
Finally, we look up the probability corresponding to the standardized value z = -1.968 in the standard normal distribution table, which will give us the probability that the sum of 121 values is less than 350.