Question

Historically, the members of the chess club have had an average height of 5

′
6
′′
with a standarid deviation of 2
′′
. What is the probability of a player being between 5
′
3
′′
and 5' 10"? (Submit your answer as a whole number. For example if you calculate 0.653 (or 65.3% ), enter 65. )
Previous question

Answer 1

The answer is 91

since, The Probability is P(5'3" < X < 5'10") = 91%

Explanation:

We assume that the heights follow a normal distribution such that we can use the z-table

Mean = M = 5'6" = 66"

Standard Deviation = S = 2"

We have to find the probability for 5'3" = 60 + 3 = 63" and 5'10" = 60 + 10 = 70" and then subtract them to get the probability of being in between.

To find the probability, we have to find the z-score and then look up the value corresponding to it to get the probability

for 5'3"

P(X < 5'3"),

Calculating the z-score,

z = (x - M)/S

z = (63 - 66)/2

z = -3/2

z = -1.5

Finding the value,

= 0.0668

So,

P(X < 5'3") = 0.0668

for 5'10"

P(X < 5'10"),

Calculating the z score,

z = (x - M)/S

z = (70 - 66)/2

z = 4/2

z = 2

And looking up the corresponding value we get,

= 0.9772

P(X < 5'10") = 0.9772

The probability in between is then,

P(5'3" < X < 5'10") = P(X < 5'10") - P(X < 5'3")

= 0.9772 - 0.0668 = 0.9104

so,

P(5'3" < X < 5'10") = 0.9104 = 91.04%

For whole number,

P(5'3" < X < 5'10") = 91%