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Answer:
(a) 50%
(b) 47.5%
(c) 2.5%
Step-by-step explanation:
According to the honest coin principle, if the random variable <em>X</em> denotes the number of heads in <em>n</em> tosses of an honest coin (<em>n</em> ≥ 30), then <em>X</em> has an approximately normal distribution with mean, and standard deviation,
.
Here the number of tosses is, <em>n</em> = 2500.
Since <em>n</em> is too large, i.e. <em>n</em> = 2500 > 30, the random variable <em>X</em> follows a normal distribution.
The mean and standard deviation are:
(a)
To not lose any money the even rolls has to be 1250 or more.
Since, <em>μ</em> = 1250 it implies that the 50th percentile is also 1250.
Thus, the probability that by the end of the evening you will not have lost any money is 50%.
(b)
If the number of "even rolls" is 1250, it implies that the percentile of 1250 is 50th.
Then for number of "even rolls" as 1300,
1300 = 1250 + 2 × 25
= μ + 2σ
Then P (μ + 2σ) for a normally distributed data is 0.975.
⇒ 1300 is at the 97.5th percentile.
Then the area between 1250 and 1300 is:
Area = 97.5% - 50%
= 47.5%
Thus, the probability that the number of "even rolls" will fall between 1250 and 1300 is 47.5%.
(c)
To win $100 or more the number of even rolls has to at least 1300.
From part (b) we now 1300 is the 97.5th percentile.
Then the probability that you will win $100 or more is:
P (Win $100 or more) = 100% - 97.5%
= 2.5%.
Thus, the probability that you will win $100 or more is 2.5%.
Answer:
Step-by-step explanation:
Given that a survey was conducted that asked 1004 people how many books they had read in the past year.
Results are:
x bar =12.5
s = 16.6
n = 1004
Std error of sample =
Since population std dev is not known t critical value can be used for confidence interval.
df = 1004-1 =1003
t critical = 1.96
Margin of error
Confidence interval lower bound
Upper bound