Assume women’s heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. a. If one wom
2 answers:
Answer: a) 0.5636 b) 0.7881
Step-by-step explanation:
We assume that women’s heights are normally distributed .
Let x be the random variable that represents the shoe sizes.
Also, The population mean = ; Standard deviation:
a) Formula for z:-
Put x= 64, we get
Now, the probability that the male shoe sizes are greater than 8 :-
Hence, the probability that her height is less than 64 inches = 0.5636
b. Sample size : n= 25
Then , the formula for z :-
Put x= 64, we get
Then, the probability that they have a mean height less than 64 inches.:_
Hence, the probability that they have a mean height less than 64 inches. =0.7881
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Answer:
(a) The probability of overbooking is 0.2135.
(b) The probability that the flight has empty seats is 0.4625.
Step-by-step explanation:
Let the random variable <em>X</em> represent the number of passengers showing up for the flight.
It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.
Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.
Number of reservations = 19
Regular customers = 14
Seats available = 17 - 14 = 3
Remaining reservations, n = 19 - 14 = 5
P (A remaining passenger will arrive), <em>p</em> = 0.52
The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.52.
(1)
Compute the probability of overbooking as follows:
P (Overbooking occurs) = P(More than 3 shows up for the flight)
Thus, the probability of overbooking is 0.2135.
(2)
Compute the probability that the flight has empty seats as follows:
P (The flight has empty seats) = P (Less than 3 shows up for the flight)
Thus, the probability that the flight has empty seats is 0.4625.