Question

Among 625 random digits, find the probability that the digit 7 appears: (a) between 50 and 60 times inclusive, (b) between 60 and 70 times inclusive.

Answer 1

Answer:

Given,

sample n = 625

Total digits = 0,1,2,3,4,5,6,7,8,9

P(Getting 7) = 1/10

np = 625*1/10

= 62.5

nq = 625*(1-1/10)

= 625*9/10

= 562.5

Here np & nq > 5, so we use normal approximation

Mean = np = 62.5

Standard deviation = sqrt(npq)

= sqrt(625*1/10*9/10)

= sqrt(56.25)

= 7.5

a)

P(50 < X < 60) = P(49.5 < X < 60.5)

= P((49.5 - 62.5)/7.5 < (x-u)/s < (60.5 - 62.5)/7.5)

= P(-1.73 < z < -0.27)

= P(z < -0.27) - P(z < -1.73)

= 0.3935801 - 0.0418151 [since from z table]

= 0.3518

b)

P(60 < X < 70) = P(59.5 < X < 70.5) [continuity correction]

= P((59.5 - 62.5)/7.5 < (x-u)/s < (70.5 - 62.5)/7.5)

= P(-0.4 < z < 1.07)

= P(z < 1.07) - P(z < -0.4)

= 0.8576903 - 0.3445783 [since from z table]

= 0.5131