Suppose that the times taken for germination for cauliflower seeds are normally distributed with a mean of 6.8 days. Suppose also that exactly $90 \%$ of the cauliflower seeds germinate in 5.9 days or more. Find the standard deviation of times taken for germination for cauliflower seeds. Carry your intermediate computations to at least four decimal places. Round your answer to at least two decimal places.

Step 1 ：We are given that the times taken for germination for cauliflower seeds are normally distributed with a mean of 6.8 days. Also, exactly 90% of the cauliflower seeds germinate in 5.9 days or more. This implies that 10% of the seeds germinate in less than 5.9 days. In a standard normal distribution, the value that leaves 10% in the left tail corresponds to a z-score of -1.2816.

Step 2 ：We can use this information to set up the following equation and solve for the standard deviation (σ): \(z = \frac{X - μ}{σ}\), where z is the z-score, X is the value from the original distribution (5.9 days), μ is the mean (6.8 days), and σ is the standard deviation.

Step 3 ：Substituting the given values into the equation, we get: \(-1.2816 = \frac{5.9 - 6.8}{σ}\).

Step 4 ：Solving this equation for σ, we find that σ is approximately 0.7022. This is the standard deviation of the times taken for germination for cauliflower seeds.

Step 5 ：However, the problem asks us to round our answer to at least two decimal places. So, the rounded standard deviation is 0.70.

Step 6 ：Final Answer: The standard deviation of the times taken for germination for cauliflower seeds is approximately \(\boxed{0.70}\) days.