Final answer:
The probability that the sum of the lengths of five randomly chosen tiles exceeds 99.4 cm can be found using the normal distribution properties. We calculate the mean and standard deviation of the sum and use the z-score to find the corresponding probability from the z-table.
Step-by-step explanation:
To find the probability that the sum of the lengths of five randomly chosen tiles exceeds 99.4 cm, we need to use the properties of the normal distribution. Since the length of each tile is normally distributed with a mean (μ) of 19.8 cm and a standard deviation (σ) of 0.1 cm, the sum of the lengths of five tiles is also normally distributed. The mean of the sum is 5 × 19.8 cm = 99 cm, and the standard deviation of the sum is √5 × 0.1 cm = √0.5 cm.
The probability question can be answered by finding the z-score and using the standard normal distribution. To calculate the z-score, the formula used is:
Z = (X - μtotal)/σtotal
Where X = 99.4 cm, μtotal = 99 cm, and σtotal = √0.5 cm. Plugging in the values:
Z = (99.4 - 99)/√0.5 = 0.4/√0.5
Now, this z-score can be used to find the corresponding probability from the z-table, which gives us the required probability.