Mathematics, 17.03.2020 19:30 iamasia06

Assume that females have pulse rates that are normally distributed with a mean of mu equals 72.0 beats per minute and a standard deviation of sigma equals 12.5 beats per minute. Complete parts (a) through (c) below.
a. If 1 adult female is randomly selected, find the probability that her pulse rate is between 66 beats per minute and 78 beats per minute.
The probability is?

b. If 4 adult females are randomly selected, find the probability that they have pulse rates with a mean between 66 beats per minute and 78 beats per minute
The probability is?

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?
A.
Since the mean pulse rate exceeds 30, the distribution of sample means is a normal distribution for any sample size.
B.
Since the distribution is of sample means, not individuals, the distribution is a normal distribution for any sample size.
C.
Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.
D.
Since the distribution is of individuals, not sample means, the distribution is a normal distribution for any sample size.

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Answer from: uday50

a) 36.88% probability that her pulse rate is between 66 beats per minute and 78 beats per minute

b) 66.30% probability that they have pulse rates with a mean between 66 beats per minute and 78 beats per minute

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

To solve this question, we 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 sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 72, \sigma = 6.5$

a. If 1 adult female is randomly selected, find the probability that her pulse rate is between 66 beats per minute and 78 beats per minute.

The probability is?

This is the pvalue of Z when X = 78 subtracted by the pvalue of Z when X = 66. So

X = 78

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

$Z = \frac{78 - 72}{12.5}$

Z = 0.48

Z = 0.48 has a pvalue of 0.6844

X = 66

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

$Z = \frac{66 - 72}{12.5}$

Z = -0.48

Z = -0.48 has a pvalue of 0.3156

0.6844 - 0.3156 = 0.3688

36.88% probability that her pulse rate is between 66 beats per minute and 78 beats per minute

b. If 4 adult females are randomly selected, find the probability that they have pulse rates with a mean between 66 beats per minute and 78 beats per minute

The probability is?

Now we have $n = 4, s = \frac{12.5}{\sqrt{4}} = 6.25$

X = 78

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

By the Central Limit Theorem

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

$Z = \frac{78 - 72}{6.25}$

Z = 0.96

Z = 0.96 has a pvalue of 0.8315

X = 66

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

$Z = \frac{66 - 72}{6.25}$

Z = -0.96

Z = -0.96 has a pvalue of 0.1685

0.8315 - 0.1685 = 0.6630

66.30% probability that they have pulse rates with a mean between 66 beats per minute and 78 beats per minute

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

The condition for the sample size exceeding 30 is when the population is skewed. If it is normally distributed, the size is not a condition.

So the correct answer is:

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Answer from: Quest

Num of plants = (original#ofplants)*(1.6)^t

Answer from: Quest

215.92 dollars is the answer

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