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Numerade Educator



Problem 63 Medium Difficulty

The weights of ripe watermelons grown at Mr. Smith's farm are normally distributed with a standard deviation of 2.8 lb. Find the mean weight of Mr. Smith's ripe watermelons if only $3 \%$ weigh less than 15 lb.




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Video Transcript

okay for this problem. We're giving some watermelon information, and we told it's normally distributed. But what's interesting about this problem is that we don't know the mean for it, so we know we have watermelon distribution, and we know the standard deviation for it. So the standard deviation for that is plus or minus £2.8 for these ripe watermelons we want to find out is we want to find out what the means for is now the most useful in piece of information here is we're, um we know that 3% way less than £15. So we actually do know this £15 mark over here and 3% way less than that. So our plan for this problem is going to be, too. Since we can figure out the area that corresponds to 3% you can find a Z score for the 3% and then we can figure out the meat weight is So I use my calculator over here, and it was called the Universe Norms that we're gonna inverse when work backwards off of our known information can figure out with the area that corresponds to 3% is and we know the area poor corresponds to 3%. We're gonna find out the value that goes with 3%. And so we know that 3% value goes with the Z score of negative 1.8 somebody's e 15 here because what that meant was his £15 apples. Since we knew the apple information for this problem, I mean watermelon, not at all. So that's negative. 1.88 So our next step here, the first step was to get that Z, that's a Z score. Their money there is the ah formula to figure out the unknown mean problem there. So we know any of these four equals two. The number we know. So I was gonna go attacks here. There. We know my Aniston. I mean, divided by the standard deviation of this problem. Really, we're trying to find is the the mean So let's put in the values we know negative 1.88 And we know the negative 1.88 forest corresponds to the 15 there, minus the mule, divided by the 2.8 standard deviations for this data. Someone who's my calculator now. Ah, So if I understand a little bit of algebra to get this alone. I need to multiply both sides by 2.8. So I'm just gonna hit the times, and that will give me we have over here in the left side of my equation. Equals to 15 minus meal. Okay. Is it? Technically, I would add the mean to the right side of the left side of the equation and then add the 5.2662 figure with them. You is. So let's just do this here, Um, you eagles to whatever. The 15 plus 2 5.26 sixes. So I'm gonna take that answer. And I wasn't going to be tricky there, but I messed myself up. Welcome news. I wrote it down. Okay, so 5.266 is what I need to do. So, any 5.266 need Adebowale sides? Really? What that means is I'm at about 2 15 and so that's going to be my final meet. Answer. Uh, which is 20 points to six. So 22.6 is the mean weight of the watermelons. The unknown mean Wait