Answer:
a) μ = 99 σ = 3.333
b) μ = 99 σ = 2.673
c) μ = 99 σ = 1.69
d) μ = 99 σ = 1.291
e) μ = 99 σ = 0.913
f) μ = 99 σ = 0.456
Explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation
![s = (\sigma)/(√(n))](https://img.qammunity.org/2021/formulas/mathematics/college/tqgdkkovwzq5bzn3f9492laup3ofuhe2qd.png)
In this problem, we have that:
![\mu = 99, \sigma = 10](https://img.qammunity.org/2021/formulas/mathematics/college/uxe8pk0dbovcos3g2aeafqk337bf176q2n.png)
(a) n=9 μ = σ =
![\mu = 99, s = (10)/(√(9)) = 3.333](https://img.qammunity.org/2021/formulas/mathematics/college/1v7eesva7mnrthj6o0gxrnwtywaajepa4c.png)
(b) n=14 μ = σ =
![\mu = 99, s = (10)/(√(14)) = 2.673](https://img.qammunity.org/2021/formulas/mathematics/college/yr0ahlcr2r2vo00d6p5dmocuwk4tx4elf2.png)
(c) n=35 μ = σ =
![\mu = 99, s = (10)/(√(35)) = 1.69](https://img.qammunity.org/2021/formulas/mathematics/college/5fem6q9picqvv1vsojfk1g4z3cnu2yhugw.png)
(d) n=60 μ = σ =
![\mu = 99, s = (10)/(√(60)) = 1.291](https://img.qammunity.org/2021/formulas/mathematics/college/ag9dr7iavpfwn6lmw05p79curpd2oxje54.png)
(e) n=120 μ = σ =
![\mu = 99, s = (10)/(√(120)) = 0.913](https://img.qammunity.org/2021/formulas/mathematics/college/wcbfvm6x835wy61gpgpo7tzc1tacrrcgc7.png)
(f) n=480 μ = σ =
![\mu = 99, s = (10)/(√(480)) = 0.456](https://img.qammunity.org/2021/formulas/mathematics/college/qixwj7pymfyp4vhb0ffb1vum70kusge5y3.png)