Question

The length of human pregnancies from conception to birth is normally distributed with mean 266 days and standard deviation 16 days. What is the proportion of the lengths of pregnancies that fall between 250 days and 282 days?

Please put the answer in the standard deviation percentages!

Answer 1

Answer:

68% of the lengths of pregnancies fall between 250 days and 282 days.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 266

Standard deviation = 16.

What is the proportion of the lengths of pregnancies that fall between 250 days and 282 days?

250 = 266 - 16

So 250 is one standard deviation below the mean.

282 = 266 + 16

So 282 is one standard deviation above the mean.

By the Empirical Rule, 68% of the lengths of pregnancies fall between 250 days and 282 days.

Answer 2

Answer:

$P(250<X<282)=P(\frac{250-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{282-\mu}{\sigma})=P(\frac{250-266}{16}<Z<\frac{282-266}{16})=P(-1<z<1)$

And we can find this probability with this difference and using the normal standard table or excel:

$P(-1<z<1)=P(z<1)-P(z<-1)= 0.8413-0.1587= 0.6826$

So then we will have approximatetly 68.26% of the values between 250 and 282 days

Step-by-step explanation:

Let X the random variable that represent the The length of human pregnancies from conception to birth, and for this case we know the distribution for X is given by:

$X \sim N(266,16)$  

Where $\mu=266$ and $\sigma=16$

We are interested on this probability

$P(250<X<282)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(250<X<282)=P(\frac{250-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{282-\mu}{\sigma})=P(\frac{250-266}{16}<Z<\frac{282-266}{16})=P(-1<z<1)$

And we can find this probability with this difference and using the normal standard table or excel:

$P(-1<z<1)=P(z<1)-P(z<-1)= 0.8413-0.1587= 0.6826$

So then we will have approximatetly 68.26% of the values between 250 and 282 days