in a blind esp test, a person correctly identifies whether a tossed coin comes up head or tails in 63 trials out of 100. Using the normal approximation (without the continuity correction), which of the following would you use go calculate the probability of correctly identifying 63 or more?
A. 1-binomcdf(200, .5, 62)
B. P (z-5.233)
C. P (x63)
D. P(z-.0707
I think it's C,but it says 63 or more and C doesn't include 63 so i'm confused? please help

Question

in a blind esp test, a person correctly identifies whether a tossed coin comes up head or tails in 63 trials out of 100. Using the normal approximation (without the continuity correction), which of the following would you use go calculate the probability of correctly identifying 63 or more?
A. 1-binomcdf(200, .5, 62)
B. P (z>-5.233)
C. P (x>63)
D. P(z>-.0707
I think it's C,but it says 63 or more and C doesn't include 63 so i'm confused? please help

kropot72 · Answer

Using the normal approximation to the binomial distribution, the mean number of successes is np = 100 * 0.63 = 63. Can you now see what the probability of correctly identifying 63 or more must be? This should enable you to confirm the correct choice.

anonymous · Answer

so would it be C then or would I use one of the z-scores to correctly identify the probability of 63 or more? @kropot72

kropot72 · Answer

The z-score for 63 is zero. P(63 or more successes) = 05.

kropot72 · Answer

Also, P(63 or fewer successes) = 0.5.

anonymous · Answer

so it has to be either A or C and since the total is 100 and not 200 it can't be A so it has to be C, Thank you!

kropot72 · Answer

Good work! You're welcome :)