We are given that the lengths of pregnancies in a small rural village are normally distributed with a mean of 262 days and a standard deviation of 17 days.

Let X = lengths of pregnancies in a small rural village

SO, X ~ Normal( $\mu=262,\sigma^{2} = 17^{2}$ )

Here, $\mu$ = population mean = 262 days

$\sigma$ = standard deviation = 17 days

Now, the 68-95-99.7 rule states that;

68% of the data values lies within one standard deviation points.

95% of the data values lies within two standard deviation points.

99.7% of the data values lies within three standard deviation points.

So, middle 68% of most pregnancies is represented through the range of within one standard deviation points, that is;

[ $\mu -\sigma$ , $\mu + \sigma$ ] = [262 - 17 , 262 + 17]

= [245 days , 279 days]

Hence, the range in which we can expect to find the middle 68% of most pregnancies is [245 days , 279 days].

5 0

3 years ago

The radius of a circle is 50 feet what is the circumstances of the circle use 3.14 for pi

oksian1 [2.3K]

Answer:

314 feet

Step-by-step explanation:

$c = 2\pi \: r \\ = 2 \times 3.14 \times 50 \\ = 3.14 \times 100 \\ = 314 \: feet$

6 0

3 years ago

A mountain climber started at zero elevation. The first day he hiked 10 miles up

Mandarinka [93]

Answer:

14 miles if that's how far he went up

17miles if you're calculating his distance covered

4 0

2 years ago