. Given a standard normal distribution, answer the following questions a. Find t
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Question
. Given a standard normal distribution, answer the following questions a. Find the probability that x2.16 b. Find the probability that-1.05rs2 c. Find the probability that ?-122 d. Find the value of a such that (zaa)-0.5698 2. Birth weight in the U.S.A. is normally distributed with a mean of 3560 gram and a standard deviation of 435 gram. a. Find the probability that any randomly selected baby has weight less than 2960 grams b. Find the probability that any randomly selected baby has weight no less than 3100 grams c. Find the probability that any randomly selected baby has weight between 2960 grams and 3000 grams d. In a sample of 30 randomly selected babies, what is the probability that their average weights is no more than 3300 grams e. If American Medical Association considers that the lightest 5% babies needs the special observation, find the cut off weight of babies who need special observation. 3. Assume 20 % of all Students will plan to study abroad in next year and we randomly select 50 students a. Find the probability that in the sample we select, more than 10 of them plan to study abroad in next year b. Find the probability that in the sample we select, exact 10 of them plan to study abroad in next year. c. Find the probability that in the sample we select, less than half of them plan to study abroad in next year. d. Find the average number of student in this sample will plan to study abroad in next year and also find its standard deviation. e. In this sample, is it usual to have less than 15 of them to study abroad in next year in nest year? State your reason. 4. Use the given the table below to answer the following questions 0.52 0.33 0.05 0.10 a. Is the given table a probability distribution table? State the reason b. If it is, find its mean and standard deviationExplanation / Answer
Question 1:
a) From the standard normal tables, we get:
P(Z < 2.16 ) = 0.9846
Therefore 0.9846 is the required probability here.
b) P( -1.05 < Z < 2 )
= P(Z < 2 ) - P(Z < -1.05 )
Getting it from the standard normal tables, we get here:
= 0.9772 - 0.1469
= 0.8303
Therefore 0.8303 is the required probability here.
c) P(Z > -1.22 )
= 1 - P( Z < -1.22 )
Getting it from the standard normal tables, we get:
= 1 - 0.1112
= 0.8888
Therefore 0.8888 is the required probability here.
d) P(Z > a ) = 0.5698
Therefore, P(Z < a ) = 1 - 0.5698 = 0.4302
From standard normal tables, we have:
P(Z < -0.1759 ) = 0.4302
Therefore a = - 0.1759
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