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:
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