mark noticed that the probability that a certain player hits a home run in a single game is 0.175. mark is interested in the variability of the number of home runs if this player plays 200 games. if mark uses the normal approximation of the binomial distribution to model the number of home runs, what is the standard deviation for a total of 200 games? answer choices are rounded to the hundredths place.



Answer :

The standard deviation for a total of 200 games with p is 0.175 and q is 0.825 is 5.37.

The likelihood of receiving one of two outcomes under a set of conditions is modeled by the probability distribution known as the binomial distribution. The variance formula for this distribution is npq, and the standard deviation is √(npq). Here, p is the success probability, n is the number of trials, and q is the failure probability.

Given that p is 0.175 and n is 200. Then, q = 1-0.175 = 0.825

Then, the variance is

[tex]\begin{aligned}\sigma^2&=200\times0.175\times0.825\\&=28.875\end{aligned}[/tex]

The standard deviation will be

[tex]\begin{aligned}\sigma&=\sqrt{npq}\\&=\sqrt{28.875}\\&=5.37\end{aligned}[/tex]

The required answer is 5.37.

To know more about binomial distribution:

https://brainly.com/question/14565246

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