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Problem 51 Hard Difficulty

Depending on where you live and on the quality of the day care, costs of day care can range from $\$ 3000$ to $\$ 15,000$ a year (or $\$ 250$ to $\$ 1250$ a month) for one child, according to the Baby Center. Day care centers in large cities such as New York and San Francisco are notoriously expensive. Suppose that day care costs are normally distributed with a mean equal to $\$ 9000$ and a standard deviation equal to $\$ 1800 .$
a. What percentage of day care centers cost between $\$ 7200$ and $\$ 10,800 ?$
b. What percentage of day care centers cost between $\$ 5400$ and $\$ 12,600 ?$
c. What percentage of day care centers cost between $\$ 3600$ and $\$ 14,400 ?$
d. Compare the results in a through c with the empirical rule. Explain the relationship.

Answer

a .6826; b .9545; c .9973 d. +/- 1 SD, 2 SD, and 3 SD for a, b, and c, respectively

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Video Transcript

in problem number 51. They're giving us some information about the daycare costs, um, that are normally distributed. The day care costs have a mean of $9000 per year, and we have a standard deviation off $1800. And there are four parts to this problem. And in part one, it's asking us for a percentage of daycare centers that cost between 7200 and 10,800. So, in order to find that percentage were first gonna find the probability and then merge was going to switch that into a percentage. And in order to accomplish this, we are going to start with the bell shaped curve. And on that bell shaped curve, we're gonna put the $9000 average. And in this problem, we're talking about being between 7200 and 10,800 so we will need the scores and to refresh your memory on the formula for a Z score, it's X minus mu over sigma, So the Z score associated with 7200 would be 7200 minus 9000, divided by 1800. And with that calculation, you will get negative one so we can put a negative one above the 7200. And the Z score associated with the 10,800 would be 10,800 minus 9000 divided by 1800 and you would get a positive one so we can put a positive one up on our picture as well. So if you think about what the question is actually asking again, it says, What is the percentage? And we're doing Probability was probability that we have day care costs between 217,800. It's no different than saying What's the probability that is? The score is between negative one and positive one. So we're going to calculate the probability that a Z scores less than one. And from that we're going to subtract the probability that is the scores less than negative one. And we're going to do that by utilizing your standard normal table in the back of your book. And if you look up Z being less than positive one, you're going to get a 10.8 413 and the Z being less than negative one will be 0.1587 So you get a probability of 0.68 26 so that transitions into 68.26% of daycares will have a cost between $217,800 in a year. Now let's go to Part B and in part B, you are asked to determine again, Ah, percentage. But we will first find a probability before we do the percentage. And we want to know between 5400 and 12,600. So we're going to do the probability that the X scores between 5400 and 12,600 we're gonna construct are bell shaped curve again. Again, we have 9000 in the center, and this time we want to be between 5400 in 12,600. So we're going to calculate each of their disease scores. So Z equals 5400 minus 9000, divided by 1800 and you will get a negative, too. And then the second Z's score for this will be 12,600 minus 9000, divided by 1800 and you'll get a positive, too. So negative two is associated with 5400 positive. Two is associated with 12,600. Someone were asking for the percentage with a probability, between 417,600. It's no different than saying What's the probability that your Z score is between negative two and positive, too? So to solve that will have to find the probability that C is less than positive, too. And from that attract the probability that Z is less than negative, too. We're going to utilize our table in the back of the book and the probability that Z is less than positive two would be 20.9773 And the probability that Z is less than negative to would be 0.0 to 28 for an overall probability of 0.9545 So when I transition that into a percentage were saying that 95.45% of daycares we'll have a cost of between 417,600 let's transition to part C in part C. We want to know the percentage of daycare centers that costs between 3600 and 14,400. Again, we have our normal distribution with 9000 in the center, 3600 is to the left, and 14,400 is to the right, and we will need the Z score associated with each one. So the Z score, associated with 3600 will be calculated 3600 minus 9000 divided by 1800 which is a negative three, and Z equals 14,400 minus 9000, divided by 1800 and you get a positive three. So when were saying What's the probability that we will find daycare centers that have a cost between 3600 and 14,400? It's no different than saying What's the probability that your Z score is between negative three and positive three? So we're gonna do the pot probability that Z is less than positive three. And from that we're gonna subtract the probability that C is less than negative. Three. We're going to look in the back of the book and for positive three, you're going to find an area of 30.9987 and for negative three, you're going to find an area of 30.13 And when you subtract those, you're gonna get 0.9974 Which means that 99.74% of daycares will fall between $3600.14,400 dollars now for Part D in part, Dietz talking us about or talking to us about the empirical rule and it says, compare the results in a through C with the empirical rule so very quickly. Let's just recap what the probabilities were for a so in part a our probability or our percentage was 68.26%. In part B R probability was 95.45% and in part, C R probability was 99 0.74%. So now we want to look at the empirical rule, and the empirical rule states that within one standard deviation, So when we go between negative one is Aziz score and positive One is a C score, you will find approximately 68% of the data. So we came up with 68.26% Empirical rule approximates at 68% if you are within two standard deviations of the mean meaning between a Z score of negative two and positive, too. The empirical rule accounts for 95% and we had an answer of 95.45%. So within two, standard deviations should be 95 or approximately 95%. And when we did our calculations between negative to a positive two, we got close to that. The final part of the empirical rule says when you are within three standard deviations of the means. So that means between the Z score of negative three and positive three, it will account for 99.7% of the data and in our part C, we got 99.74%. So keep in mind the empirical rule is Justin approximate. There they are in Prague, approximate values, and it is more accurate to use your Z scores and your standard normal tables. If you want to carry it out to more decimal places and be more precise