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A school knows that 11% of its students are left-handed. What is the probability that there will be between 33 and 45 left-handed students in a randomly selected group of 400? Use your calculator and this formula: P(a≤x≤b) = binomcdf(n,p,b) - binomcdf(n,p,a - 1). Round your answer to three decimal places.

A. 0.029
B. 0.043
C. 0.573
D. 0.602​

Respuesta :

Answer:0.573

Step-by-step explanation:

MrRoyal

The probability that there will be between 33 and 45 left-handed students in a randomly selected group of 400 is 0.573

How to determine the probability?

The given parameters are:

  • Sample size, n = 400
  • Probability of success, p = 0.11
  • x = 33 to 45

To determine the required probability, we make use of:

P(33 ≤ x ≤ 45) = P(x ≤ 45) - P(x < 33)

Using a binomial calculator, we have:

P(33 ≤ x ≤ 45) = 0.60226197339 - 0.02904256227

Evaluate the difference

P(33 ≤ x ≤ 45) = 0.57321941112

Approximate

P(33 ≤ x ≤ 45) = 0.573

Hence, the probability that there will be between 33 and 45 left-handed students in a randomly selected group of 400 is 0.573

Read more about binomial probability at:

https://brainly.com/question/15246027

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