Question

Heights of 10 year olds. Heights of 10 year olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches. Round all answers to two decimal places.

1. What is the probability that a randomly chosen 10 year old is shorter than 67 inches?

2. What is the probability that a randomly chosen 10 year old is between 60 and 63 inches?

3. If the tallest 15% of the class is considered very short, what is the height cutoff for very short?

4. What is the height of a 10 year old who is at the 65 th percentile?

Answer 1

Given,

$\mu$ = 55, $\sigma$ = 6

We convert this to standard normal as

P( X < x) = P( Z < x - $\mu$ / $\sigma$ )

a )

P( X < 67) = P( Z < 67 - 55 / 6)

= P( Z < 2)

= 0.9772

P( X < 67) = 0.98

b)

P( 60 < X < 63) = P( X < 63) - P( X < 60)

= P( Z < 63 - 55 / 6) - P( Z < 60 - 55 / 6)

= P( Z < 1.3333) - P( Z < 0.8333)

= 0.9088 - 0.7977

= 0.1111

P( 60 < X < 63) = 0.11

3)

We have to calculate x such that P( X > x) = 0.15

That is

P( X < x) = 0.85

P( Z < x - $\mu$ / $\sigma$ ) = 0.85

From the Z table, z-score for the probability of 0.85 is 1.0365

x - $\mu$ / $\sigma$ = 1.0365

Put the values of $\mu$ and $\sigma$ in above expression and solve for x

x - 55 / 6 = 1.0365

x = 61.22

4)

p^th perecentile = $\mu$ + Z $\sigma$ , Where Z is critical value at give confidence level.

so,

65th percentile = 55 + 0.3853 * 6

= 57.31