Question

Suppose babies born in a large hospital have a mean weight of 3242 grams, and a standard deviation of 446 grams. If 107 babies are sampled at random from the hospital, what is the probability that the mean weight of the sample babies would differ from the population mean by less than 40 grams? Round your answer to four decimal places.

Answer 1

Answer: 0.3524

Step-by-step explanation:

Given : Mean : $\mu = 3242\text{ grams}$

Standard deviation :  $\sigma = 446\text{ grams}$

Sample size : $n= 107$

To find the probability that the mean weight of the sample babies would differ from the population mean by less than 40 grams i.e. more than 3202 grams and less than 3282 grams

The formula for z-score :-

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For x = 3202 grams

$z=\dfrac{3202-3242}{\dfrac{446}{\sqrt{107}}}=-0.93$

For x = 3282 grams

$z=\dfrac{3282-3242}{\dfrac{446}{\sqrt{107}}}=0.93$

The P-value= $P(-0.93<z<0.93)=2P(z<-0.93)=2(0.1761855)=0.352371\approx0.3524$

Hence, the required probability =0.3524