The correct option e) 94.29%, for the probability that among the next 166 responses there will be at most 43 correct answers.
If we use a normal distribution to simulate any binomial distribution with p = 0.24 for n = 166 results, then:
There should be 43 correct responses, or
μ = np = 166*0.24
μ = np = 34.86
The number of accurate responses has a standard deviation of
σ = √(np(1-p))
σ = √(34.86*0.76)
σ = 5.14
43 correct responses have a z-value;
z = (x - μ)/σ
z = (43 - 34.86)/5.14
z = 1.58
The likelihood that 43 accurate responses are provided is:
N(1.58) = 0.9429
Thus, the probability that among the next 166 responses there will be at most 43 correct answers is 94.29%.
To know more about the Normal approximation, here
https://brainly.com/question/28194998
#SPJ4
The complete question is-
Use the normal distribution to approximate the desired probability. A certain question on a test is answered correctly by 24.0 percent of the respondents. Estimate the probability that among the next 166 responses there will be at most 43 correct answers.
a) 74.56870% b) 74.30203% c) 75.20203% d) 94.95% e) 94.29%