Question

The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with μ = 110 grams and σ = 25 grams. A sample of 25 vitamins is to be selected.

(a) What is the probability that the sample mean will be between 100 and 120 grams?

(b) What is the probability that the sample mean will be less than 100 grams?

(c) 95% of all sample means will be greater than how many grams?

(d) The middle 70% of all sample means will fall between what two values?

Answer 1

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Params of normal dist:

mean = 110
Stdev = 25
n = 25

a.

P(100<x<120) = P(100 - 110 / sqrt(25) <Z< (120-110)/sqrt(25)) = P(-2<Z<2) = .95

b.

P(X<100) = P( Z < 100-110 / sqrt(25)) = P(Z<-2) = .025

c.

95% of all sample means will be greater than , lets say c
So, P(X>c) = .95
P(Z> (c-110)/(25/sqrt(25))) = .95
(c-110)/5 = -1.645
c = 110+ 5*-1.645 = 101.775

So, 95% of all sample means will be greater than 101.775

d.

Middle 70% is. Let middle 70% be designated by a Z scores of x and y

So, P(x<Z<y) = .70 or P(Z<x) = .15

So, So, (x-110)/(25/sqrt(25)) = -1.036433389

x = 110+ 5*-1.036433389 = 104.82
y = 110 + 5*1.036433389 = 115.18

So, 70% is between 104.82 and 115.18