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stellarik [79]
3 years ago
6

Hey 17% discount on a watch saved a shopper $51. Find the price of the watch before the discount

Mathematics
2 answers:
konstantin123 [22]3 years ago
7 0
Discount =17%=51 $(given)
Let the price be=x
So formula is
Discount%/100 *actual price =discount (in$)
17/100*x=51
X=51*100/17
X=300$
Hence the price of the watch is 300$
Nookie1986 [14]3 years ago
6 0
 <span>To rephrase the question,
17% of what is $51?

Let me give you a short way or formula.
is/of = %/100

is = 51
0f = x
%= 17%

51/x = 17/100
51*100 = x*17
5100= 17x
5100/17= x

so x = 300</span>
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c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

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To solve this question, we need to understand the normal probability distribution and the central limit theorem.

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Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean of a population is 74 and the standard deviation is 15.

This means that \mu = 74, \sigma = 15

Question a:

Sample of 36 means that n = 36, s = \frac{15}{\sqrt{36}} = 2.5

This probability is 1 subtracted by the pvalue of Z when X = 78. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{78 - 74}{2.5}

Z = 1.6

Z = 1.6 has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

Question b:

Sample of 150 means that n = 150, s = \frac{15}{\sqrt{150}} = 1.2247

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

Z = \frac{X - \mu}{s}

Z = \frac{77 - 74}{1.2274}

Z = 2.45

Z = 2.45 has a pvalue of 0.9929

X = 71

Z = \frac{X - \mu}{s}

Z = \frac{71 - 74}{1.2274}

Z = -2.45

Z = -2.45 has a pvalue of 0.0071

0.9929 - 0.0071 = 0.9858

0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c. A random sample of size 219 yielding a sample mean of less than 74.2

Sample size of 219 means that n = 219, s = \frac{15}{\sqrt{219}} = 1.0136

This probability is the pvalue of Z when X = 74.2. So

Z = \frac{X - \mu}{s}

Z = \frac{74.2 - 74}{1.0136}

Z = 0.2

Z = 0.2 has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

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