Question

1) The household income in a certain community is normally distributed with a mean of $42,000 and a standard deviation of $5,000. The proportion of households with incomes of at least $50000 is between: a) 5% and 6% b) 44% and 45% c) 94% and 95% d) none

2) The actual weight of "8 oz. chocolate bar" produced by a certain machine are normally distributed with mean 8.1 oz. and standard deviation of 0.1 oz. only 5% of the bars weigh less than approximately:

a) 8.2645oz

b) 7.9355 oz

c) 8.000 oz.

d) none

Answer 1

Solution :

Given that ,

mean = = 42000

standard deviation = = 5000

1)

P(x 50000) = 1 - P(x   50000)

= 1 - P[(x - ) / (50000-42000) / 5000]

= 1 -  P(z 1.6)   

  Using z table,

= 1 - 0.9452

=0.0548

a.between 5% and 6%

2)

Solution :

Given that ,

mean = = 8.1

standard deviation = =0.1

Using standard normal table,

P(Z < z) = 5%

= P(Z < z) = 0.05  

= P(Z < -1.645) = 0.05

z = -1.645

Using z-score formula  

x= z * +

x= -1.645*0.1+8.1

x= 7.9355 oz