Question

Some IQ tests are standardized to a Normal distribution with mean 100 and standard deviation 16. (a) [5 points] What proportion of IQ scores is between 95 and 105? (b) [5 points] What is the 80th percentile of the IQ scores? (c) [5 points] A random sample of 10 candidates are about to take the test. What is the probability that at least half of them will score between 95 and 105? (d) [5 points] A random sample of 10 candidates are about to take the test. What is the probability that their average score will be between 95 and 105?

Answer 1

Here = 100, = 16

a) Here we need to find

p ( 95 < x < 105 )

= p ( -0.31 < z < 0.31 )

= p ( z < 0.31 ) - p ( z < -0.31 )

= 0.6217 - 0.3783

= 0.2434

b) 80th Percentile :

p ( Z z ) = 0.80

z =0.84

x =113.44

So 80th percentile is 113.44.

c ) Here we have random sample of size 10. i.e. n = 10 , p ( score between 95 and 105 ) = 0.2434

We apply here binomial distribution .

P (x =x ) = ⁿC_x * p ^x * ( 1 - p ) ^n-x

We need to find  the probability that at least half of them will score between 95 and 105.

p ( x 5 ) = p ( x = 5) + p ( x = 6) + p(X = 7 ) + p ( x = 8 ) + p ( x =9 ) + p ( x = 10 )

=¹⁰C₅ * (0.2434) ⁵ * ( 1 - 0.2434)^10-5 +  ¹⁰C₆ * (0.2434) ⁶ * ( 1 - 0.2434)^10-6 + ¹⁰C₇ * (0.2434) ⁷ * ( 1 - 0.2434)^10-7

+ ¹⁰C₈ * (0.2434) ⁸ * ( 1 - 0.2434)^10-8 + ¹⁰C₉ * (0.2434) ⁹ * ( 1 - 0.2434)^10-9 + ¹⁰C₁₀ * (0.2434) ¹⁰ * ( 1 - 0.2434)^10-10  

= 0.0707

d) Here n = 10

We need to find, the probability that their average score will be between 95 and 105.

( 95 < < 105 )

= p ( -0.99 < z < 0.99 )

= p ( z < 0.99 ) - p ( z < -0.99 )

= 0.8389 - 0.1611

= 0.6778