Question

In: Statistics and Probability

A research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes...

A research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. The population mean is $25 and the population standard deviation is $15. What is the probability that a sample of $100 steady smokers spend between $24 and $26?

a. 0.495

b. 0.252

c. 0.248

d. 0.748

Solutions

Expert Solution

Solution :

Given that ,

mean =   = 25

standard deviation = = 15

n = 100

= 25

=  / n= 15/ 100=1.5

P(24<     <26 ) = P[(24-25) / 1.5< ( - ) /   < (26-25) /1.5 )]

= P(-0.67 < Z < 0.67)

= P(Z <0.67 ) - P(Z <-0.67 )

Using z table

=0.7486 - 0.2514

=0.495

probability= 0.495

orchestra answered 1 year ago
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