The heights of the trees in a forest are normally distributed, with a mean of 25 meters and a standard deviation of 6 meters. Wh
2 answers:
Since roughly 95% of a normal distribution lies within two standard deviations of the mean, you know that about 5% lies without, and since the distribution is symmetric, you can expect about 2.5% of the distribution to lie above two standard deviations from the mean. So the probability is about
If you want more precision, the actual value is closer to
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Answer:
(1) Correct option (A).
(2) The probability that Aadi will get Tails is 0.40.
Step-by-step explanation:
The information provided is:
- Eric throws a biased coin 10 times. He gets 3 tails.
- Sue throw the same coin 50 times. She gets 20 tails.
The probability of tail in both cases is:
Here,
P (T|E) implies the probability of tail in case of Eric's experiment.
P (T|S) implies the probability of tail in case of Sue's experiment.
(1)
Now, it is given that Aadi is going to throw the coin once.
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
In this case we need to compute the probability of Aadi getting Tails in a single toss.
As Sue uses a larger number of trials in the experiment, i.e. n = 50 > 30 times, according to the Central limit theorem, Sue's estimate is best because she throws it .
Thus, the correct option is (A).
(2)
As explained in the first part that Sue's estimate is best for getting a tail, the probability that Aadi will get Tails when he tosses the coin once is:
Thus, the probability that Aadi will get Tails is 0.40.
Let X be the number of female employee. Let n be the sample size, p be the probability that selected employee is female.
It is given that 45% employee are female it mean p=0.45
Sample size n=60
From given information X follows Binomial distribution with n=50 and p=0.45
For large value of n the Binomial distribution approximates to Normal distribution.
Let p be the proportion of female employee in the given sample.
Then distribution of proportion P is normal with parameters
mean =p and standard deviation =
Here we have p=0.45
So mean = p = 0.45 and
standard deviation =
standard deviation = 0.0642
Now probability that sample proportions of female lies between 0.40 and 0.55 is
P(0.40 < P < 0.45) =
= P(-0.7788 < Z < 1.5576)
= P(Z < 1.5576) - P(Z < -0.7788)
= P(Z < 1.56) - P(Z < -0.78)
= 0.9406 - 0.2177
= 0.7229
The probability that the sample proportion of females is between 0.40 and 0.55 is 0.7229
