Answer:
305/4
Step-by-step explanation:
Answer: I think it’s A
Step-by-step explanation:
Hope this helps
Answer:
The range in which we can expect to find the middle 68% of most pregnancies is [245 days , 279 days].
Step-by-step explanation:
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 = <u><em>lengths of pregnancies in a small rural village</em></u>
SO, X ~ Normal(
)
Here,
= population mean = 262 days
= standard deviation = 17 days
<u>Now, the 68-95-99.7 rule states that;</u>
- 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;
[
,
] = [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].
Step-by-step explanation:
The total number of people mentioned in this problem is 15. 15 will be the denominator for the probability in fraction form. The sum of the # of people that include lifeguards and managers will be the numerator.
Therefore, the probability that the person picked is either a manager or a lifeguard is 10/15 [because there are 2 managers and 8 lifeguards; 2 + 8 =10].
<u>Fraction form of probability: 10/15</u>
<u>Decimal form: 0.6 repeated</u>
<u>Percent form: about 66.67%</u>