N_s = sample size = 900. μ_s = mean of the sample = 3.4 cm

σ_s = Standard deviation of the sample = 2.61 cm

Population mean μ₀ =3.25

Population standard deviation = σ₀ = 2.61 cm

*student's t = (μ_s - μ₀) / [σ_s / √N_s ] *

* t * = (3.4 - 3.25) * √900 / 2.61 **= 1.7241**

Find
the probability that -1.7241 <= t <= 1.7241 from the
students t distribution table or from a website that calculates these.
Here the degrees of freedom (d.f) are 899 = N_s - 1. Sample size is 900.

The probability we get is 0.915. Hence, we can say that the sample belongs to the
population with a probability 0.915 or * a confidence level of 91.5%.*

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**The population is normal. we don't know its mean. We know sample size, mean and standard deviation.** We are given confidence level 95% = 0.95

Normal distribution variable:

Z = (μ_s - μ₀) / (σ_s/√N_s)

Here σ_s = 2.61 cm, N_s = 900

We consult the Standard Normal Distribution tables. We find z such that

P(0 <= Z <= z ) = 0.95/2 = 0.475,

or from the cumulative standard distribution table we find z such that :

P(-∞ < Z <= z ) = 0.5 + 0.95/2 = 0.975

Either way, we get the value z of variable Z for which the probability

P( -z <= Z <= z) = 95% = 0.95 is z = 1.96

-1.96 <= Z <= 1.96

-1.96 * σ_s <= √N_s * (μ_s - μ₀) <= 1.96 * σ_s

- 0.1705 <= (μ_s - μ₀) <= 0.1705 cm

As, μ_s = 3.4 cm :

* 3.4 - 0.1705 <= μ₀ <= 3.4+0.1705 cm*

ie., range [3.23 cm , 3.57 ]

That will be the range of the population mean that we will estimate from the available sample data.