Question

11.

A research firm used a sample of 200 teenagers to estimate the mean amount teenagers spend on junk food during a week. The population mean was known to be $25 with a standard deviation of $6.

a) Calculate the standard error. [1]

b) What is the probability that a sample of 200 teenagers spend between $24.00 and $25.50?

[2]

c) What is the probability that teenagers spend at least $24.50? [2]

Answer 1

Solution :

Given that,

mean = = $ 25

standard deviation = = $ 6

n = 200

a) = = $ 25

= / n = 6 / 200 = 0.424

b) P(24.00 < < 25.50)  

= P[(24.00 - 25) / 0.424 < ( - ) / < (25.50 - 25) / 0.424)]

= P(-2.36 < Z < 1.18)

= P(Z < 1.18) - P(Z < -2.36)

Using z table,  

= 0.8810 - 0.0091

= 0.8719

c) P( 24.50) = 1 - P( 24.50)

= 1 - P[( - ) / (24.50 - 25) / 0.424 ]

= 1 - P(z -1.18)

Using z table,    

= 1 - 0.1190

= 0.8810