Answer:
78.74% of the 787-8 airplanes are between 185' and 187'.
Explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2021/formulas/mathematics/college/c62rrp8olhnzeelpux1qvr89ehugd6fm1f.png)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
![\mu = 185.4, \sigma = 0.5](https://img.qammunity.org/2021/formulas/mathematics/college/234vvq41ty9gcs8rbk7w3pqxtzkoi8n1ho.png)
What proportion of the 787-8 airplanes are between 185' and 187'?
This is the pvalue of Z when X = 187 subtracted by the pvalue of Z when X = 185. So
X = 187
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2021/formulas/mathematics/college/c62rrp8olhnzeelpux1qvr89ehugd6fm1f.png)
![Z = (187 - 185.4)/(0.5)](https://img.qammunity.org/2021/formulas/mathematics/college/s40ulf7r0gfo5eskaatvif1u0735d404ts.png)
![Z = 3.2](https://img.qammunity.org/2021/formulas/mathematics/college/8v6zju7dfmjfahd90cbx1k9i5qqwr1bldz.png)
has a pvalue of 0.9993.
X = 185
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2021/formulas/mathematics/college/c62rrp8olhnzeelpux1qvr89ehugd6fm1f.png)
![Z = (185 - 185.4)/(0.5)](https://img.qammunity.org/2021/formulas/mathematics/college/tgvoa89hlqp1gvpp59attuc0u3d499sqgc.png)
![Z = -0.8](https://img.qammunity.org/2021/formulas/mathematics/college/kkhnpkiefckabwmly12vsk03f6b1bmr4zt.png)
has a pvalue of 0.2119.
0.9993 - 0.2119 = 0.7874
78.74% of the 787-8 airplanes are between 185' and 187'.