Question

questions 3 – 4, assume that an airplane has an entry door with a height of 70 in. Assume heights of adult women are normally distributed with a mean of 65 in. and a standard deviation of 2.5 in.

3. What doorway height would allow 98% of adult women to fit without bending? Round to the nearest inch.

4. If 30 women are randomly selected and board the airplane, what is the probability that their mean height will be less than 70 inches? Round to the nearest percentage point (for example, 52%).

Answer 1

Hello Sir/ Mam

Let X be the random variable that shows the height of adult women.

Hence, X is normally distributed with a mean of 65 inches and standard deviation of 2.5 inches.

Q - 3 -

We have to calculate X such that Probability is 98%.

Hence,

We an easily calculate the z-score for 98% probability using excel formula, "=NORMSINV(98%)"

We get that z-score = 2.0537

Hence,

Hence, the required height is 70.13 inches.

Q - 4 -

Hence, the required probability :

I hope this solves your doubt.

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