Question

A commuter study in Hobart on September 1st, 2019 found that the cost of travelling to work varied according to a normal distribution with a mean of $4.98 per day and a standard deviation of 70 cents per day.

1. What is the probability that a randomly selected commuter paid less than $5.91 per day? (4dp)

2. What is the probability that a randomly selected commuter paid more than $4.68 per day? (4dp)

3. What is the probability that a randomly selected commuter paid between $4.05 and $5.54 per day? (4dp)

4. If the cost to a commuter was on the 60th percentile, what was the cost of travelling to work? (2dp) $

Answer 1

Solution :

Given that ,

mean = = 4.98

standard deviation = = 0.70

a)

P(x <5.91 ) = P((x - ) / < (5.91 - 4.98) / 0.70)

= P(z < 1.33)

= 0.9082 Using standard normal table,

Probability = 0.9082

b)

P(x > 4.68) = 1 - P(x < 4.68)

= 1 - P((x - ) / < (4.68 - 4.98) / 0.70)

= 1 - P(z < -0.43)

= 1 - 0.3336 Using standard normal table.

= 0.6664

Probability = 0.6664

c)

P(4.05 < x < 5.54) = P((4.05 - 4.98)/ 0.70) < (x - ) /  < (5.54 - 4.98) / 0.70) )

= P(-1.33 < z < 0.8)

= P(z < 0.8) - P(z < -1.33)

= 0.7881 - 0.0918 Using standard normal table,  

Probability = 0.6963

d)

P( Z < z ) = 60%

P( Z < z ) = 0.60

P( Z < 0.253 ) = 0.60

z = 0.253

Using z - score formula,

X = z * +

= 0.253 * 0.70 + 4.98  

= 5.16