Question

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 66 inches?

2. What is the probability that a randomly chosen 10 year old is between 59 and 64 inches?

3. If the shortest 5% of the class is considered very tall, what is the height cutoff for very tall?   inches

4. What is the height of a 10 year old who is at the 36 th percentile?

Answer 1

Solution :

Given that ,

mean = = 55

standard deviation = = 6

1)

P(x < 66) = P((x - ) / < (66 - 55) / 6)

= P(z < 1.83)

Using standard normal table,

P(x < 66) = 0.9664

Probability = 0.9664 = 0.97

2)

P(59 < x < 64) = P((59 - 55)/ 6) < (x - ) / < (64 - 55) / 6) )

= P(0.67 < z < 1.5)

= P(z < 1.5) - P(z < 0.67)

= 0.9332 - 0.7486 = 0.1846

Probability = 0.18

3)

P(Z < z) = 5% = 0.05

P(Z < -1.645) = 0.05

z = -1.645

Using z-score formula,

x = z * +

x = -1.645 * 6 + 55 = 45.13

Height cutoff = 45.13

4)

P(Z < z) = 36%

P(Z < -0.36) = 0.36

z = -0.36

Using z-score formula,

x = z * +

x = -0.36 * 6 + 55 = 52.84

36th percentile = 52.84