Question

(b) The time, Y minutes,

taken by Sid Slow to install a satellite dish may also be assumed to be
a normal random variable, but with
P(Y< 170) = 0.14 and P(y >200) = 0.03.
Determine, to the nearest minute, values for the mean and standard deviation of Y.

Answer 1

mean = 181

standard deviation = 10

Explanation:

Here, we want to calculate the values for the mean and standard deviation of y

Mathematically;

z-score = (x-mean)/SD

where x is the score we are looking at

Let the mean be a and the standard deviation be b

Firstly, we need the z-scores that correlate to the individual probabilities

For P(Y < 170) = 0.14

We can use the standard normal table for this;

The z-score here = -1.08

For P (y > 200) ;

The z-score here = 1.881

So for the first case:

-1.08 = (170 - a)/b

-1.08b = 170 - a

a - 1.08b = 170 •••••••••(i)

For the second case

1.881 = (200 - a)/b

1.881b = 200 - a

a + 1.881b = 200 ••••••• (ii)

So we have two equations to solve simultaneously;

a - 1.08b = 170

a + 1.881b = 200

Subtract equation ii from i

1.881b + 1.08b = 200-170

2.961b = 30

b = 30/2.961

b = 10.13

But a - 1.08b = 170

a = 170 + 1.08b

a = 170 + 1.08(10.13)

a = 170 + 10.94

a = 180.94

To the nearest minutes;

a = mean = 181

b = standard deviation = 10