If a baseball player's batting average is 0.340 (i. e., the probability of getting a hit each time at bat is 0.340), find the probability that the player will have a bad season and get at most 60 hits in 200 times at bat

13.1%

Step-by-step explanation:

Given:

N = 200  

P= 0.34

0<= x <=60

x= number of hits.

n=total times at bat

In order to find mean, we take the product of n and p, therefore:

mean 'μ' = np = 200 x  0.34 = 68

next is to find standard deviation, i.e the square root of the product of n, p and q

where q is 1-p

SD 'σ' = sqrt(npq) = sqrt(68*0.66) = 6.6993  

applying continuity correction,

z = (x - μ) / σ    

z = (60 - 68) /6.6993 = - 1.12

For the normal standard distribution,

P(z < - 1.12) = 13.14% => 13.1%

the answer is d. (1.2, -4.7) on edgn

hope this !

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step-by-step explanation: