Assume that females have pulse rates that are normally distributed with a mean of mu equals 73.0μ=73.0 beats per minute and a standard deviation of sigma equals 12.5σ=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 76 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 76 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 mean pulse rate exceeds 30, the distribution of sample means 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 distribution is of sample means, not individuals, the distribution is a normal distribution for any sample size.

D.

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

a) 0.59483

b) 0.88493

c) Option D is correct.

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:

This is a normal distribution problem with

μ = mean = 73 beats per minute

σ = standard deviation = 12.5 beats per minute

The probability that a female's pulse rate is less than 76 beats per minute. P(x < 76)

To do this, we first normalize/standardize the 76 beats per minute

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (76 - 73)/12.5 = 0.24

To determine the probability that a female's pulse rate is less than 76 beats per minute. P(x < 76) = P(z < 0.24)

We'll use data from the normal probability table for these probabilities

P(x < 76) = P(z < 0.24) = 0.59483

b) If 25 adult females are randomly selected, the probability that they have pulse rates with a mean less than 76 beats per minute.

This is still a normal distribution problem

The mean of the sample = The population mean

μₓ = μ = 73 beats per minute

But the standard deviation of the sample is related to the standard deviation of the population through the relation

σₓ = σ/√n

where n = Sample size = 25

σₓ = 12.5/√25

σₓ = 2.5

we then normalize/standardize the 76 beats per minute

z = (x - μ)/σ = (76 - 73)/2.5 = 1.20

To determine the probability that the mean of 25 female's pulse rate is less than 76 beats per minute. P(x < 76) = P(z < 1.20)

We'll use data from the normal probability table for these probabilities

P(x < 76) = P(z < 1.20) = 0.88493

c) The problem was to find the probability thay the sample mean is less than a value.

For a normal distribution, the distribution of sample mean too, is normal. It is as simple as that.

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

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