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Unformatted text preview: 95.25% of households in the city have annual incomes greater than
$30,000.
( ) ( ) ( )
( ) Therefore we expect 0.5934×60≈36 households in the sample to have annual
incomes between $35,000 and $45,000.
3. What is the 25th percentile of the normal distribution N(10, 9)? Hint: Find
the value, z, of a standardized normal random variable, Z, that is such that
(
)
and then ‘unstandardize’ to get the equivalent value for a
N(10,9) variate.
Let x be the required percentile. First find z, the 25th percentile of a standard
normal.
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( )
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) ( ) 4. In a certain city it is estimated that 40% of households have access to the
internet. A company wishing to sell services to internet users randomly
chooses 150 households in the city and sends them advertising material. For
the households contacted:
(a) Calculate the probability that less than 60 households have internet
access?
Let X be the number of households contacted that have internet access. Then
assume X is a binomial rand...
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 DenzilGFiebig

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