Question

Find the probability that in 200 tosses of a fair​ die, we will obtain at most 30 fives. Use the normal distribution to approximate the desired probability. Round to four decimal places. A. 0.4936 B. 0.3229 C. 0.1871 D. 0.2954

Answer 1

Mean = n * p

= 200 * 1/6

= 33.3333

Standard deviation = Sqrt (n p (1 -p ) )

= sqrt( 200 * 1/6 * 5/6)

= 5.27046

Using normal approximation with continuity correction,

P( X <= x) = P( Z < x+0.5 - mean / SD)

So,

P( X <= 30) = P( Z < 30.5 - 33.3333 / 5.27046 )

= P( Z < -0.5376)

= 1 - P( Z < 0.5376)

= 1 - 0.7046 (From Z table)

= 0.2954