According to one survey, the mean serum cholesterol level for U.S. adults was 203.5 , with...

According to one survey, the mean serum cholesterol level for U.S. adults was 203.5 , with a standard deviation of 43.6. simple random sample of 119 adults is chosen. Find the 46 percentile for the sample mean.

Write only a number as your answer . Round to one decimal places (for example : 204.1) . Do not write any units .

Solutions

Expert Solution

Solution:-

Given that,

mean = = 203.5

standard deviation = = 43.6

n = 119

= 203.5

= / n = 43.6 / 119 = 3.9968

The z - distribution of the 46% is ,

P(Z < z) = 46%

= P(Z < z ) = 0.46

= P(Z < -0.100 ) = 0.46

z = -0.100

Using z-score formula

= z * +

= -0.100 * 3.9968+203.5

= 203.1

Answer = 203.1

orchestra answered 1 year ago

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