Mark noticed that the probability that a certain player hits a home run in a single game is 0.165. Mark is interested in the variability of the number of home runs if this player plays 150 games. If Mark uses the normal approximation of the binomial distribution to model the number of home runs, what is the variance for a total of 150 games? Answer choices are rounded to the hundredths place.

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Priatouri Priatouri · Answer 1 · 2020-07-14T05:11:32+02:00

Answer:

Variance for a total of 150 games is 20.67

Step-by-step explanation:

If x follows binomial distribution with parameters n and p then the probability distribution of [tex]\frac{x-np}{\sqrt{npq}}[/tex] tends to N (0.1)

As [tex]n\rightarrow \infty[/tex] where E(x)= np, var (x) = npq

[tex]\therefore[/tex] var (x) = 150[tex]\times[/tex]0.165 [tex]\times[/tex] (1-0.165)

(since, n= 150, p=0.165,

q = 1-p = 1-0.165)

= 20.66625

var (x) = 20.67

[tex]\therefore\\[/tex] variance for a total of 150 games is 20.67

nishantwork777 nishantwork777 · Answer 2 · 2021-12-31T13:50:55+01:00

Given to us:

Mark uses the normal approximation of the binomial distribution to model the number of home runs. Therefore,

Probability, p=0.165;

Number of samples, n=150;

Chances of failure, q = 1-p = 1-0.165 = 0.835;

According to binomial distribution,

Variance, var(x) = npq;

var (x) = npq

= 150 x 0.165 x 0.835

= 20.66625

Hence, the variance for a total of 150 games is 20.66625.

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