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3 years ago

the sum of two numbers is 52. If one number is 76 times the second number, then find the largest number out of the two.

Mathematics

1 answer:

Leona [35]3 years ago

5 0

Answer:

The larger number is 51.3. (to 3 s.f.)

Step-by-step explanation:

Let the smaller number be x and the larger number be y.

Given that the sum of the 2 numbers is 52,

x +y= 52 -----(1)

One number is 76 times the second number. Since we have made y to be the larger number,

y= 76x -----(2)

Substitute equation (2) into (1):

x +76x= 52

Simplify:

77x= 52

Divide both sides by 77:

x= 52 ÷77

$x = \frac{52}{77}$

x= 0.675 (3 s.f.)

Substitute the value of x into (2):

$y = 76( \frac{52}{77} )$

$y = 51 \frac{25}{77}$

y= 51.3 (3 s.f.)

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a) By the Central Limit Theorem, the distribution would be approximately normal, with mean $\mu = 0.45$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.45*0.55}{100}} = 0.0497$

b) 4.02 standard deviations below the mean.

c) Making 25 or less shots has a pvalue of -4.02. Z-scores lower than -2 are considered surprising, so yes, it would be surprising for him to make only 25 of these shots.

Step-by-step explanation:

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

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

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

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

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.

For proportions, we have that the mean is $\mu = p$ and the standard deviation is $s = \sqrt{\frac{p(1-p)}{n}}$

In this problem, we have that:

$p = 0.45, n = 100$

a)By the Central Limit Theorem, the distribution would be approximately normal, with mean $\mu = 0.45$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.45*0.55}{100}} = 0.0497$

b)This is Z when X = 0.25.

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

By the Central Limit Theorem

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