Question

According to a government study among adults in the 25- to
34-year age group, the mean amount spent per year on reading and
entertainment is $1,896. Assume that the distribution of the
amounts spent follows the normal distribution with a standard
deviation of $596. **(Round your z-score computation
to 2 decimal places and final answers to 2 decimal
places.)**

What percent of the adults spend more than $2,350 per year on reading and entertainment?

What percent spend between $2,350 and $2,800 per year on reading and entertainment?

What percent spend less than $1,150 per year on reading and entertainment?

Answer #1

Solution :

Given that ,

mean = = 1896

standard deviation = = 596

a) P(x >2350) = 1 - p( x< 2350 )

=1- p P[(x - ) / < (2350-1896) /596 ]

=1- P(z < 0.76)

= 1 - 0.7764 = 0.2236

PROBABILITY = 0.2236

Answer = 22.36%

b)

P( 2350< x <2800 ) = P[(2350-1896)/596 ) < (x - ) / < (2800-1896) /596 ) ]

= P(0.76 < z <1.52 )

= P(z < 1.52) - P(z <0.76 )

=0.9357 -0.7764

PROBABILITY = 0.1594

Answer = 15.94%

c)

P(x < 1150 ) = P[(x - ) / < (1150-1896) /596 ]

= P(z <-1.25 )

= 0.1056

Probability = 0.1056

Answer =10.56 %

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