Question

For a sample of 25 observations of a normal distribution with mean 18 and
standard deviation 4.8, for 95% confidence level, calculate:
a) Lower limit of the interval for the sample mean:
b) Upper limit of the interval for the sample mean:
c) What is the probability of finding an average between 15 and 20: 

STEP BY STEP IF POSSIBLE

Answer 1

a)

sample mean, xbar = 18
sample standard deviation, s = 4.8
sample size, n = 25
degrees of freedom, df = n - 1 = 24

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.064

ME = tc * s/sqrt(n)
ME = 2.064 * 4.8/sqrt(25)
ME = 1.981

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (18 - 2.064 * 4.8/sqrt(25) , 18 + 2.064 * 4.8/sqrt(25))
CI = (16.02 , 19.98)

Lower limit = 16.02

b)

Upper limit = 19.98
c)

Here, μ = 18, σ = 0.96, x1 = 15 and x2 = 20. We need to compute P(15<= X <= 20). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (15 - 18)/0.96 = -3.13
z2 = (20 - 18)/0.96 = 2.08

Therefore, we get
P(15 <= X <= 20) = P((20 - 18)/0.96) <= z <= (20 - 18)/0.96)
= P(-3.13 <= z <= 2.08) = P(z <= 2.08) - P(z <= -3.13)
= 0.9812 - 0.0009
= 0.9803