Bags of Baby Carrots The weights of bags of baby carrots are n...

Question

Bags of Baby Carrots The weights of bags of baby carrots are normally distributed, with a mean of 32 ounces and a standard deviation of 0.36 ounce. Bags in the upper 4.5% are too heavy and must be repackaged. What is the most a bag of baby carrots can weigh and not need to be repackaged?

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says the life spans in months of a certain type of rechargeable battery are normally distributed with a mean mu equal to 24 months, and a standard deviation sigma equal to 1.2 months. Batteries in the top 4.5% of life spans are considered too long lasting for a particular device and must be replaced with different models. What is the maximum lifespan a battery can have without needing to be replaced? And we're given four possible choices as our answers. For choice A, we have 25.2 months, for choice B, we have 26.0 months, for choice C, we have 27.9 months, and for choice C, we have 26.6 months. So we're asked to find the maximum life span a battery can have without needing to be replaced. And we're also told that the batteries in the top 4.5% of life spans are considered too long lasting. So that means that we're looking for a battery life span that is greater than 95.5% of all batteries. So that would be the 95.5th percentile. And so we need to do is to find the Z score for that percentile. So, the cumulative probability for that percentile will be P of Z is less than lowercase z. Which is going to be equal to, and here we need to convert the 95.5% into a decibel by divided by 100. This is going to be equal to 0.955. So we look up the corresponding Z score in our standard normal distribution tables, and we see that our Z score Z is going to be approximately equal to. 1.70. So now what we want to do is convert this Z score into an actual lifespan. And for this, we'll need to recall a formula for calculating a Z score. So recall that that is going to be Z is equal to the quantity of X minus mu in quantity divided by sigma. For Z here is our Z score, X is going to be our lifespan, mu is going to be our mean, and sigma is going to be our standard deviation. So we'll go ahead and substitute in the values that we were given in the problem. And here, for Z we'll use our calculated Z value of 1.70, and that's going to be equal to the quantity of X, which is the value that we're looking for, minus mu, and we're told that our mean is going to be 24 months, so we'll have X minus 24, and that is going to be divided by The standard deviation, which we're told the problem is 1.2 months. So, we'll divide this by 1.2. And so now we just need to solve this equation for X. So the first thing we want to do is to multiply both sides of the equation by 1.2 to remove the fraction. So doing so, we will get 2.04. is equal to X minus 24. And to solve for X, we'll just need to add 24 to both sides, and so we'll have X is equal to 26.0. So here we just rounded to one decimal place. And for the units of this, it's going to be month. That's because both Are mean and standard deviation head units of months. So our final answer is going to be 26.0 month. And if we compare our answer here to our answer choices, we see the correct answer choice corresponds to answer choice B. So, I hope that this explanation has been useful, and I'll see you in the next video.

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