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:
Answers
Answered by
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
Similar questions