PQ = 3x + 14 and QR = 7x

Airplanes approaching the runway for landing are required to stay within the localizer (a certain distance left and right of the

Arisa [49]

Answer:

a) P(x = 0) = 64.69%

b) P(x ≥ 1) = 35.31%

c) E(x) = 0.42

d) var(x) = 0.3906

Step-by-step explanation:

The given problem can be solved using binomial distribution since:

There are n repeated trials independent of each other.
There are only two possibilities: exceedence happens or exceedence doesn't happen.
The probability of success does not change with trial to trial.

The binomial distribution is given by

P(x) = ⁿCₓ pˣ (1 - p)ⁿ⁻ˣ

Where n is the number of trials, x is the variable of interest and p is the probability of success.

For the given scenario. the six daily arrivals are the number of trials

Number of trials = n = 6

The probability of success = 7% = 0.07

a) Find the probability that on one day no planes have an exceedence.

Here we have x = 0, n = 6 and p = 0.07

P(x = 0) = ⁶C₀(0.07⁰)(1 - 0.07)⁶⁻⁰

P(x = 0) = (1)(0.07⁰)(0.93)⁶

P(x = 0) = 0.6469

P(x = 0) = 64.69%

b) Find the probability that at least 1 plane exceeds the localizer.

The probability that at least 1 plane exceeds the localizer is given by

P(x ≥ 1) = 1 - P(x < 1)

But we know that P(x < 1) = P(x = 0) so,

P(x ≥ 1) = 1 - P(x = 0)

We have already calculated P(x = 0) in part (a)

P(x ≥ 1) = 1 - 0.6469

P(x ≥ 1) = 0.3531

P(x ≥ 1) = 35.31%

c) What is the expected number of planes to exceed the localizer on any given day?

The expected number of planes to exceed the localizer is given by

E(x) = n×p

Where n is the number of trials and p is the probability of success

E(x) = 6×0.07

E(x) = 0.42

Therefore, the expected number of planes to exceed the localizer on any given day is 0.42

d) What is the variance for the number of planes to exceed the localizer on any given day?

The variance for the number of planes to exceed the localizer is given by

var(x) = n×p×q

Where n is the number of trials and p is the probability of success and q is the probability of failure.

var(x) = 6×0.07×(1 - 0.07)

var(x) = 6×0.07×(0.93)

var(x) = 0.3906

Therefore, the variance for the number of planes to exceed the localizer on any given day is 0.3906.

6 0

3 years ago

Which fraction equals the ratio of rise to run between the points (-3,-4) and (0, 0)?

Marizza181 [45]

Answer:

4/3

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distr

Marianna [84]

Answer:

(a) Decision rule for 0.01 significance level is that we will reject our null hypothesis if the test statistics does not lie between t = -2.651 and t = 2.651.

(b) The value of t test statistics is 1.890.

(c) We conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) P-value of the test statistics is 0.0662.

Step-by-step explanation:

We are given that a recent study focused on the number of times men and women who live alone buy take-out dinner in a month.

Also, following information is given below;

Statistic : Men Women

The sample mean : 24.51 22.69

Sample standard deviation : 4.48 3.86

Sample size : 35 40

Let $\mu_1$ = mean number of times men order take-out dinners in a month.

$\mu_2$ = mean number of times women order take-out dinners in a month

(a) So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in the mean number of times men and women order take-out dinners in a month}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2\neq$ 0 {means that there is difference in the mean number of times men and women order take-out dinners in a month}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_-_n_2_-_2$

where, $\bar X_1$ = sample mean for men = 24.51

$\bar X_2$ = sample mean for women = 22.69

$s_1$ = sample standard deviation for men = 4.48

$s_2$ = sample standard deviation for women = 3.86

$n_1$ = sample of men = 35

$n_2$ = sample of women = 40

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 4.48^{2}+(40-1)\times 3.86^{2} }{35+40-2} }$ = 4.16

So, test statistics = $\frac{(24.51-22.69)-(0)}{4.16 \sqrt{\frac{1}{35}+\frac{1}{40} } }$ ~ $t_7_3$

= 1.890

(b) The value of t test statistics is 1.890.

(c) Now, at 0.01 significance level the t table gives critical values of -2.651 and 2.651 at 73 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) Now, the P-value of the test statistics is given by;

P-value = P( $t_7_3$ > 1.89) = 0.0331

So, P-value for two tailed test is = 2 $\times$ 0.0331 = 0.0662

4 0

3 years ago

A woman making $3000 per month has her salary reduced by 20% because of sluggish sales. One year later, after a dramatic improve

vichka [17]

Answer:

4%

Step-by-step explanation:

Since she first goes down 20% you would do 3000 times 20 which is 60000 and divided by 100 which is 600 and you would subtract 600 from 3000 which is 2400 and then you would multiply that by 30 which is 72000 and then you would divide by 100 which is 720 and add that to 2400 which is 3120 and to find the percent change from 3000 to 3120 you would subtract 3120 by 3000 which is 120 and you would divide that by 3000 which is 0.04 and multiply that by 100 which is 4 so the percent change would be 4%

8 0

3 years ago

What is absolute value

jeka94

Absolute value is the distance between a point and 0. Absolute value is always positive. When you're writing numbers in absolute value, make sure you add:

| |

The absolute value of a negative number is also positive. For example:

| -7 | = 7 ⇒The absolute value of -7 is 7.

3 0

3 years ago

PQ = 3x + 14 and QR = 7x - 10; Find x.