Answer:
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.
