Statistics and probability

Question

The age of children in kindergarten on the first day of school is uniformly distributed between 4.86 and 5.78 years old. A first time kindergarten child is selected at random. Round answers to 4 decimal places if possible.

The mean of this distribution is

The standard deviation is

The probability that the the child will be older than 5 years old?

The probability that the child will be between 5.06 and 5.46 years old is

If such a child is at the 76th percentile, how old is that child?

Answer 1

mean = (4.86 + 5.78)/2 = 5.32

standard deviation = sqrt((5.78-4.86)^2/12) = 0.2656

P( > 5) = (5-4.86)/(5.78-4.86) = 0.152

0.76 = (x - 4.86)/(5.78-4.86) =

x = 5.5592 (76th percentile is probability of 0.76)

Answer 2

a = 4.86

b = 5.78

mean = (a+b)/2 = 5.32

Std dev = sqrt((b-a)^2)/12) = 0.2656

Use this formula to solve the next two: P(c<X<d) = (d-c)*(1/(b-a))

Older than 5: P(5<X<5.78) = 0.8478

Between 5.06 and 5.46: P(5.06 < X < 5.46) = 0.4348

76th percentile

0.76 = (x-4.86)/(5.78 - 4.86)

Solve for x

x = 5.5592