Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If 100 women are randomly selected, find the probability that they have a mean height greater than 63.0 inches.
Answers
Answered by
13
Step-by-step explanation:
Given that mean,
μ
= 63.6, standard deviation,
σ
= 2.5 and sample size,
n
= 100
The respective Z-score with sample mean,
¯
X
= 63 is
Z
=
¯
X
−
μ
σ
/
√
n
Z
=
63
−
63.6
2.5
/
√
100
Z
= -2.4
Now, we use normal tables to find the probability greater than -2.4
P (Z > -2.4) = 1 - 0.0082
P (Z > -2.4) = 0.9918
Answered by
0
Answer:
P (Z > -2.4) = 1 - 0.0082
P (Z > -2.4) = 0.9918
Step-by-step explanation:
Given that mean,
μ = '63.6' , standard deviation,
σ= '2.5' and sample size,
n= '100'
The respective Z-score with sample mean,
¯X = '63' is
Z = '¯X−μσ/√n'
Z = 63−63.6 ×2.5/√100
Z = -2.4
Now, we use normal tables to find the probability greater than -2.4
P (Z > -2.4) = 1 - 0.0082
P (Z > -2.4) = 0.9918
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