Question: a) The gene expression value of MYO7A in AML is approximated by a normal distribution model with a mean of 5 and a standard deviation
a) The gene expression value of MYO7A in AML is approximated by a normal distribution model with a mean of 5 and a standard deviation of 0.02. If a gene expression is chosen at random: i. what is the probability that the expression of this gene is between 4.98 and 5.02? ii. what is the probability that the expression of this gene is between 4.96 and 5.04? b) Suppose a data distribution is normally distributed with the population mean = 60 and the population standard deviation = 15. To find the probability of a randomly selected value less than 45, we have to find the portion of the area under the normal curve from 45 all the way to the left. i. Whether the statement given in the question and the above graph are correct according to the situation. Please justify properly. (02 pts) ii. What if we have to find the 90% percentile of the data? (01 pts) iii. Why the scale of the graph does not show - to + iv. Find the probability of randomly selected value that makes more than 80, given the same normal distribution. v. Will finite population correction will be applied here, Justify your answer with all the proofs. vi. What is the impact of standard error
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