Question

The weight of 1000 children average 50 kg and the standard deviation is 5 kg. what is the z-score corresponds to 55?<br /><br />Solution:​

Answer 1

Answer:

E(

X

ˉ

)=μ

X

ˉ

=μ=50 kg

and a variance of

{\sigma^2 \over n}={5^2 \over 30}={5 \over 6}

n

σ

2

=

30

5

2

=

6

5

The standard deviation of the sampling distribution

\sigma_{\bar{X}}={5 \over \sqrt{30}}={\sqrt{30}\over 6}σ

X

ˉ

=

30

5

=

6

30

2)

P(48<\bar{X}<53)=P\bigg({48-50 \over 6/\sqrt{30}}<Z<{53-50 \over 6/\sqrt{30}}\bigg)=P(48<

X

ˉ

<53)=P(

6/

30

48−50

<Z<

6/

30

53−50

)=

=P\bigg(Z<{\sqrt{30} \over 2}\bigg)-\bigg(1-P\bigg(Z<{\sqrt{30} \over 3}\bigg)\bigg)\approx=P(Z<

2

30

)−(1−P(Z<

3

30

))≈

\approx0.9969-0.0339=0.9630≈0.9969−0.0339=0.9630

The number of sample that fall betweeen 48 and 53 kilograms

100\cdot0.9630=96.3\approx96100⋅0.9630=96.3≈96

Explanation:

I hope it is helpful for you