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romanna [79]
3 years ago
11

Assume that females have pulse rates that are normally distributed with a mean of mu equals 74.0 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 less than 80 beats per minute. The probability is nothing. ​(Round to four decimal places as​ needed.) b. If 25 adult females are randomly​ selected, find the probability that they have pulse rates with a mean less than 80 beats per minute. The probability is nothing. ​(Round to four decimal places as​ needed.) c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30? A. Since the distribution is of sample​ means, not​ individuals, the distribution is a normal distribution for any sample size. B. Since the distribution is of​ individuals, not sample​ means, 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 mean pulse rate exceeds​ 30, the distribution of sample means is a normal distribution for any sample size.
Mathematics
1 answer:
artcher [175]3 years ago
8 0

Answer:

Step-by-step explanation:

Let x be a random variable representing pulse rate of adult females. Since the pulse rates are normally distributed, then according to the central limit theorem,

z = (x - µ)/σ

Where

x = sample mean

µ = population mean

σ = standard deviation

From the information given,

µ = 74

σ = 12.5

a) P(x < 80)

For x = 80,

z = (80 - 74)/12.5 = 0.48

Looking at the normal distribution table, the probability corresponding to the z score is 0.6844

P(x < 80) = 0.6844

b) Since the number of samples, n = 25, then the formula would be

z = (x - µ)/(σ/√n)

z = (80 - 74)/(12.5/√25) = 2.4

Looking at the normal distribution table, the probability corresponding to the z score is 0.9918

P(x < 80) = 0.9918

c) the normal distribution can be used in part​ b because

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

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