Question 13 pts

Mathematics

1 answer:

oee [108]3 years ago

7 0

Answer:

1. 68%

2. 50%

3. 15/100

Step-by-step explanation:

Here, we want to use the empirical rule

1. % waiting between 15 and 25 minutes

From what we have in the question;

15 is 1 SD below the mean

25 is 1 SD above the mean

So practically, we want to calculate the percentage between;

1 SD below and above the mean

According to the empirical rule;

1 SD above the mean we have 34%

1 SD below, we have 34%

So between 1 SD below and above, we have

34 + 34 = 68%

2. Percentage above the mean

Mathematically, the percentage above the mean according to the empirical rule for the normal distribution is 50%

3. Probability that someone waits less than 5 minutes

Less than 5 minutes is 3 SD below the mean

That is 0.15% according to the empirical rule and the probability is 15/100

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At an airport, 76% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more

I am Lyosha [343]

Answer:

The probability is $P( X \le 4 ) = 0.0054$

Step-by-step explanation:

From the question we are told that

The percentage that are on time is p = 0.76

The sample size is n = 11

Generally the percentage that are not on time is

$q = 1- p$

$q = 1- 0.76$

$q = 0.24$

The probability that no more than 4 of them were on time is mathematically represented as

$P( X \le 4 ) = P(1 ) + P(2) + P(3) + P(4)$

=> $P( X \le 4 ) = \left n } \atop {}} \right.C_1 p^{1} q^{n- 1} + \left n } \atop {}} \right.C_2p^{2} q^{n- 2} + \left n } \atop {}} \right.C_3 p^{3} q^{n- 3} + \left n } \atop {}} \right.C_4 p^{4} q^{n- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{11- 1} + \left 11 } \atop {}} \right.C_2p^{2} q^{11- 2} + \left 11 } \atop {}} \right.C_3 p^{3} q^{11- 3} + \left 11 } \atop {}} \right.C_4 p^{4} q^{11- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{10} + \left 11 } \atop {}} \right.C_2p^{2} q^{9} + \left 11 } \atop {}} \right.C_3 p^{3} q^{8} + \left 11 } \atop {}} \right.C_4 p^{4} q^{7}$

$= \frac{11! }{ 10! 1!} (0.76)^{1} (0.24)^{10} + \frac{11!}{9! 2!} (0.76)^2 (0.24)^{9} + \frac{11!}{8! 3!} (0.76)^{3} (0.24)^{8} + \frac{11!}{7!4!} (0.76)^{4} (0.24)^{7}$