The weight of a sophisticated running shoe is normally distributed with a mean of 12 ounces
and a standard deviation of 0.5 ounces.
a. What is the probability that a shoe weighs more than 13 ounces?
b. What must the standard deviation of weight be in order for the company to state that 99.9%
of its shoes are less than 13 ounces?
Answers
Answer:
The weight of a sophisticated running shoe is normally distributed with a mean of 12 ounces and a standard deviation of 0.5 ounce. (a) What is the probability that a shoe weighs more than 13 ounces? (b) What must the standard deviation of weight be in order for the company to state that of its shoes are less than 13 ounces? (c) If the standard deviation remains at 0.5 ounce, what must the mean weight be in order for the company to state that of its shoes are less than 13 ounces?
Step-by-step explanation:
The variable is normally distributed with me and mu equals 12 and standard deviation sigma equals 120.5. Based on this information, we want to answer A through C. Below this question is challenge your understanding of the adorable variables are random variables that are normally distributed to solve. We use the information presented on the left and right on the left of information on how to map a Z score onto probability or vice versa. For CNN almost traditions on the right of information relative related to how to map a random variable X. Such as the one we have above, which is not standard normal onto the standard normal Z. You do the conversion given we're utilizing falling left and right to solve So for a what is probably exaggerated than 13 we map onto the Z score excellent over sigma. 30 minutes 12.5 equals two. That's the equivalent question is equivalent to the probabilities is greater than two which is 20.2 to 8 and be what must it be for 99.9% of X to be less than 13. So the probably the lessons, you know, equals 130.999 gives us the score of 3.1. Thus to find a quick sigma we rearrange its expression for the correct value. The stigma is x minus mu over zero equals 13. 12/3 130.1 equals 0.32. To finally and see if sigma stays the same. What must you be? This that 99.9% of x is less than 13 that we're doing the same thing as in part B, but now we're given rearranging from you some U is x minus two. Sigma equals 13 minus 3.1 times 0.5 equals 11.45.