Question

In a blind ESP test, a person correctly identifies whether a tossed coin comes

up heads or tails in 63 trials out of 200. Using the normal approximation
(without the continuity correction), which of the following would you use to
calculate the probability of correctly identifying 63 or more?

Answer 1

C

Explanation:

the probability of the outcome being greater than 63 is x≥63

Answer 2

The probability of correctly identifying 63 or more is P(Z > -5.233)

Calculating the Zscore :

• sample proportion , p = x/n
• population proportion, P = 0.5

From here. we have , p ;

• p = 63/200 = 0.315

Zscore = p - P / √P(1 - P)/n

• p - P = -0.185
• √P(1 - P)/n = √0.5(1 - 0.5)/200 = 0.035355

The Zscore becomes :

Zscore = -0.185 / 0.035355 = -5.2326

The probability of correctly identifying 63 or more is would be written as :

P(Z > Zscore ) = P(Z > -5.233)

Hence, the correct option is (D)