According to a market research firm, 54% of all residential telephone numbers in Los Angeles are unlisted. A telephone sales firm uses random digit dialing equipment that dials residential numbers at random, whether or not they are listed in the telephone directory. The firm calls 486 numbers in Los Angeles.
What is the probability that at least half the numbers dialed are unlisted? (Remember to check that you can use the normal approximation.)
Can you show how you got the answer please I don’t understand it
It uses the normal approximation to the binomial. You don’t say whether or not to use the continuity correction; I will assume not.
I don’t know what rule you’ve been given regarding whether the use of the normal approximation is okay (I have seen at least five different rules). A common one is that n.p and n(1-p) are both greater than 5 (which is easily satisfied)
"At least half the numbers" means the number unlisted is ≥ 243
So you need to compute P(X ≥ 243) where X is distributed Binomial(486, 0.54)
This is approximately the same as P(X ≥ 243) where
X is distributed Normal(486 x 0.54, 486 x 0.54 x 0.46)
the mean of the normal is 486 x 0.54 = 262.44
the variance of the normal is 486 x 0.54 x 0.46 = 120.7224
the standard deviation of the normal = √120.7224 = 10.9874
(NB the continuity correction would correspond to finding P(X ≥ 242.5) in the normal)
Under the normal approximation :
P(X ≥ 243) = P((X-262.44)/10.9874 ≥ (243 -262.44)/10.9874) = P(Z ≥ (243 -262.44)/10.9874 )
= P( Z ≥ -1.7693)
NB I also don’t know how your tables are organized!
Looking this up in my normal tables, P( Z ≥ -1.7693) = P(Z ≤ 1.7693) = 0.9615
So the probability that at least half the numbers are unlisted is about 96.2%
[If you use the continuity correction the answer is 96.52%. The exact (binomial) answer is 96.51%]