In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. It is estimated that 3.3% of the general population will live past their 90th birthday. In a graduating class of 794 high school seniors, find the following probabilities. (Round your answers to four decimal places.) (a) 15 or more will live beyond their 90th birthday 0.9898 (b) 30 or more will live beyond their 9oth birthday 0.257 X (c) between 25 and 35 will live beyond their 90th birthday 0.5964 (d) more than 40 will live beyond their 90th birthday 0.0023
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Hi how would u figure this out?
Given:
n = 794
p = 0.033
The sample size is large enough to use normal distribution.
Also np = 794×0.033 = 26.202 > 5
n(1-p) = 794 ×(1-0.033) = 767.798 > 5
Thus, we can use normal approximation to binomial distribution as all conditions are satisfied.
We approximate binomial distribution to normal with,
μ = n×p = 794×0.033 = 26.202
σ = √n×p×q = √794×0.033×(1-0.033) = 5.0336.
Part a:
The probability that 15 or more will live beyond their 90th birthday is computed as,
Thus, the probability that 15 or more will live beyond their 90th birthday is
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