Probability and statistics

Question

Females have mean weight of 78.6kg with standard deviation of 5.03 kg.

1. 20% of females weigh less than t kg. Find value of t , to 2 decimal places

2. Middle 90% of males weigh between an and b kg. Find value of an and b to nearest kg.

Answer 1

Hi George,

This problem uses the classic equation in elementary statistics:

z = (x-mu)/sigma

1. Assuming normality, question essentially asks for bottom 20% of female weights. Look in interior of Z-table for 0.20–decimal equivalent of 20%.

z (0.20) = - 0.84

mu = 78.6

sigma = 5.03

Solve classic equation for x:

-0.84 =( x - 78.6)/5.03

Multiply by 5.03 on both sides:

-4.2252 = x - 78.6

Add 78.6 to both sides:

x = 74.37 or

t = 74.37

#2.

Middle 90% means we want fifth and 95th percentiles, so check Z- table interior for both:

z(0.05) = -1.645

z(0.95) = 1.645

Now, we want both the lower and higher x values, so I’ll use an and b. Lower bound:

-1.645 = (a - 78.6)/5.03

Multiply by 5.03

-8.274 = a - 78.6

Add 78.6 to both sides:

a = 70.33

Now let’s get b:

1.645 = (b -78.6)/5.03

Multiply both sides by 5.03:

8.274 = (b - 78.6)

Add 78.6 on both sides:

b = 86.87

So:

a = 70.33

b = 86.87

I hope this helps.