Question

HSCI 390 Assignment 7 Name 1. Researchers found the mean sodium intake in men and women 60 years or older to be 2940 mg with a standard deviation of 1476 mg. Use these values for the mean and standard deviation of the US population and find the probability that a random sam 75 people from the population will have a mean: O M N = 2940 0=1476 M=75 X - M a) Less than 2450 mg okn b) Over 3100 mg c) Between 2500 and 3300 mg d) What is the standard error of the mean in this case? How can it be decreased?

Answer 1

a)

Here, μ = 2940, σ = 170.4338 and x = 2450. We need to compute P(X <= 2450). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (2450 - 2940)/170.4338 = -2.88

Therefore,
P(X <= 2450) = P(z <= (2450 - 2940)/170.4338)
= P(z <= -2.88)
= 0.0020

b)

Here, μ = 2940, σ = 170.4338 and x = 3100. We need to compute P(X >= 3100). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (3100 - 2940)/170.4338 = 0.94

Therefore,
P(X >= 3100) = P(z <= (3100 - 2940)/170.4338)
= P(z >= 0.94)
= 1 - 0.8264 = 0.1736

c)

Here, μ = 2940, σ = 170.4338, x1 = 2500 and x2 = 3300. We need to compute P(2500<= X <= 3300). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (2500 - 2940)/170.4338 = -2.58
z2 = (3300 - 2940)/170.4338 = 2.11

Therefore, we get
P(2500 <= X <= 3300) = P((3300 - 2940)/170.4338) <= z <= (3300 - 2940)/170.4338)
= P(-2.58 <= z <= 2.11) = P(z <= 2.11) - P(z <= -2.58)
= 0.9826 - 0.0049
= 0.9777

d)

std.error = s/sqrt(n)
= 1476/sqrt(75)
= 170.4338

by increasing th esampl esize it can be reduced