QUESTION IMAGE
Question
assume that females have pulse rates that are normally distributed with a mean of \\( \mu=72.0 \\) beats per minute and a standard deviation of \\( \sigma=12.5 \\) beats per minute. complete parts (a) through (b) below.
. if 1 adult female is randomly selected, find the probability that her pulse rate is less than 76 beats per minute.
the probability is \\( \square \\).
(round to four decimal places as needed.)
b. if 4 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 \\( \square \\).
(round to four decimal places as needed.)
Step1: Calculate the z - score for part (a)
The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\). Here, \(x = 76\), \(\mu=72\), and \(\sigma = 12.5\).
We want to find \(P(X\lt76)\), which is equivalent to \(P(Z\lt0.32)\) using the standard normal distribution.
Step2: Find the probability for part (a)
Using a standard normal table or a calculator with a normal - distribution function (\(P(Z\lt z)=\varPhi(z)\)), for \(z = 0.32\), \(\varPhi(0.32)=0.6255\) (using a standard normal table or a calculator like \(P(Z\lt0.32)=\text{normcdf}(-\infty,0.32)\) in a TI - 84 Plus: \(\text{normcdf}(-1000,0.32,0,1)\)).
Step3: Calculate the z - score for part (b)
The formula for the z - score of the sample mean \(\bar{x}\) is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). Here, \(\bar{x}=76\), \(\mu = 72\), \(\sigma=12.5\), and \(n = 4\).
We want to find \(P(\bar{X}\lt76)\), which is equivalent to \(P(Z\lt0.64)\)
Step4: Find the probability for part (b)
Using a standard normal table or a calculator with a normal - distribution function (\(P(Z\lt z)=\varPhi(z)\)), for \(z = 0.64\), \(\varPhi(0.64)=0.7389\) (using a standard normal table or \(\text{normcdf}(-1000,0.64,0,1)\) in a TI - 84 Plus)
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a. \(0.6255\)
b. \(0.7389\)