The heights of adult women are approximately normally distributed with a mean of 160cm160\operatorname{cm} and a standard deviation of 7cm7\operatorname{cm}. Find the height xx such that 5%5\% of women are shorter than xx.

148.5 cm ⁡ 148.5\operatorname{cm}

The heights of adult women are approximately normally distributed with a mean of 160 cm 160\operatorname{cm} and a standard deviation of 7 cm 7\operatorname{cm} . Find the height x x such that 5 % 5\% of women are shorter than x x .

So now that we've seen how to find values from given probabilities in a nonstandard normal distribution, let's take a look at our practice problem and hopefully had a chance to work it out on your own. So we're told here that the heights of adult women are approximately normally distributed, meaning that they follow sort of a bell shaped curve. Now, the mean is not zero and the sigma is not one. We actually have some real world values here. We're told that the mean is 160 centimeters and the standard deviation is seven centimeters. So I always like to start out with a little picture over here just to kind of sort of figure out how to sort of draw out the problem. But basically what we have here is we have a bell shaped curve, and our mu is equal to 116 centimeters, and the sigma, which is kind of how wide the distribution is, is equal to seven centimeters. Alright? Those are our values. Now we're going to find not a probability, but we're going to be finding a height here. We're going to be finding a height x such that five percent of women, just sort of overall, are going to be shorter than that value. Alright? So So in other words, we need to figure out a value so that everything to the left of that value is equal to about 5% of the total distribution of people. Okay? So in other words, where I go draw that line, it's probably gonna be to the left of this area of this middle over here because we're only want 5% to the left. So I'm just gonna go ahead and try to sort of figure out where I could draw that line such that when I shade it, this is gonna be representing about 5% of people. It's going to be right around over here somewhere. Alright? So I've got area, which is equal to 5%, or as a decimal place, this would be 0.05. Alright? So okay. So now that we go ahead and do that, how do we actually figure out what this x value is that translates that? We're going to be using this new equation over here, which again is really just sort of flipping around the z score equation that we've seen before. Let's go ahead and write our values. We've got mu is equal to 160, we've got r sigma is equal to seven, and we've got the area that we're interested in, or the probability that's given to us, is 0.05. Now remember, we're going to need we already have the sigma, we already have the mu to find out our x. We first have to figure out the z score that translates to or is corresponding to that area. Alright. That's how we use this x formula. So in other words, what I'm going to do is I'm going to take this area and I have to turn it into a z score, and that's where that comes in. What is the z score that's associated with only 5% of the distribution being to the left? Well, to figure this out, you can go ahead and use either a calculator or you can go ahead and figure out this using your z table. So you would just go over here to your z table, and you'd be looking for over here on the left side of this page. Whoops. Over here. And you'd be figuring out basically where do you have 0.05 in the distribution. And if you go ahead and look through some of those values, you're going to see that it's going to be somewhere right around over here, which is going to be somewhere in between those two values. If you were to go ahead and look to the ends of the columns and rows and just sort of take an average of those two numbers, we've actually seen this number over and over again. The z score that translates to this sort of this amount of probability or area to the left is going to be negative 1.645. It's going to be halfway between those two digits on the end there, those two hundredths places. Alright? So I'm just going to go ahead and do this negative 1.645, and now I can plug this into our x equation over here. So x is equal to I've got negative 1.645. We're going to multiply this by the sigma, which is seven, then we're going to add it to the mean, which is going to be 160. Alright. So we're going to add this to 160. Perfectly fine to have negative numbers inside of this. This just means that your x value is going to be to the left of 160, and that totally makes sense. If you actually go ahead and work this out, what you should find here is that x value is equal to I've got 148.5 centimeters. Alright? In other words, that's this number right over here. At basically a 148.5, there's only five percent of women of the distribution who are gonna be shorter than that. Alright? So this is gonna be our answer right over here. That's really all there is to it, folks. Let's go ahead and take a look at a couple more practice problems, and thanks for watching.

The heights of adult women are approximately normally distributed with a mean of 160 cm ⁡ 160\operatorname{cm} and a standard deviation of 7 cm ⁡ 7\operatorname{cm} . Find the height x x such that 5 % 5\% of women are shorter than x x .

148.5 cm ⁡ 148.5\operatorname{cm}

145.2 cm ⁡ 145.2\operatorname{cm}

151.3 cm ⁡ 151.3\operatorname{cm}

160.0 cm ⁡ 160.0\operatorname{cm}

6. Normal Distribution and Continuous Random Variables

Non-Standard Normal Distribution

Statistics

6. Normal Distribution and Continuous Random Variables

Non-Standard Normal Distribution

Struggling with Statistics?

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Multiple Choice

The heights of adult women are approximately normally distributed with a mean of $160 cm$ and a standard deviation of $7 cm$ . Find the height $x$ such that $5 percent sign$ of women are shorter than $x$ .

$145.2 cm$

$148.5 cm$

$151.3 cm$

$160.0 cm$

Verified step by step guidance

Step 1: Recognize that the problem involves a normal distribution with a mean (μ) of 160 cm and a standard deviation (σ) of 7 cm. The goal is to find the height (x) such that 5% of women are shorter than this height. This corresponds to finding the value of x for which the cumulative probability (area under the curve) is 0.05.

Step 2: Use the z-score formula to relate the height (x) to the standard normal distribution. The formula is:

z = (x - μ) / σ

, where z is the z-score, μ is the mean, and σ is the standard deviation.

Step 3: Look up the z-score that corresponds to a cumulative probability of 0.05 in a standard normal distribution table (or use statistical software). This z-score will be negative because it is below the mean. For a cumulative probability of 0.05, the z-score is approximately

z = -1.645

Step 4: Rearrange the z-score formula to solve for x:

x = μ + zσ

. Substitute the known values: μ = 160, z = -1.645, and σ = 7.

Step 5: Perform the calculation to find the value of x. This will give the height such that 5% of women are shorter than this value. Ensure the result matches one of the provided options.

3:17m

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