Mathematics, 19.06.2020 04:57, thereal6ix9ine69

. a. If 1 adult female is randomly selected, find the probability that her pulse rate is greater than 70 beats per minute. b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean greater than 70 beats per minute. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30

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answered: jenistha123

a) 34.46% probability that her pulse rate is greater than 70 beats per minute.

b) 2.28% probability that they have pulse rates with a mean greater than 70 beats per minute.

c) Because the underlying distribution(female's pulse rate) is normal.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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.

Distribution of females pulse rates:

Here, i suppose there was a typing mistake, since the mean and the standard deviation are lacking.

Also, the question c. only makes sense if the distribution is normal, so i will treat it as being.

I will use $\mu = 68, \sigma = 5$ . I am guessing these values, just using them to explain the question.

a. If 1 adult female is randomly selected, find the probability that her pulse rate is greater than 70 beats per minute.

This is 1 subtracted by the pvalue of Z when X = 70. So

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

$Z = \frac{70 - 68}{5}$

Z = 0.4

Z = 0.4 has a pvalue of 0.6554

1 - 0.6554 = 0.3446

34.46% probability that her pulse rate is greater than 70 beats per minute.

b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean greater than 70 beats per minute.

Now $n = 25, s = \frac{5}{\sqrt{25}} = 1$

This probability is 1 subtracted by the pvalue of Z when X = 70. So

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

By the Central Limit Theorem

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

$Z = \frac{70 - 68}{1}$

Z = 2

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability that they have pulse rates with a mean greater than 70 beats per minute.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30

The sample size only has to exceed 30 if the underlying distribution is normal. Here, the distribution of females' pulse rate is normal, so this requirement does not apply.

answered: Guest

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1