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The auto parts department of an automotive dealership sends out a mean of 4.34.3 special orders daily. What is the probability t

Svetlanka [38]

Answer:

$P(X = 5) = 0.166$

Step-by-step explanation:

Given

$Mean = 4.3$

$x = 5$

Required

Determine the probability that the order is 5

This question can be answered using Poisson distribution.

$P(X = x) = \frac{\alpha ^x e^{-\alpha}}{x!}$

Where

$\alpha$ is used to represent the mean

and

$\alpha = 4.3$

$x = 5$

$e = 2.71828$ ---- Euler's constant

So, we have:

$P(X = x) = \frac{\alpha ^x e^{-\alpha}}{x!}$

$P(X = 5) = \frac{4.3^5 * 2.71828^{-4.3}}{5!}$

$P(X = 5) = \frac{4.3^5 * 2.71828^{-4.3}}{5* 4 * 3 * 2 *1}$

$P(X = 5) = \frac{4.3^5 * 2.71828^{-4.3}}{120}$

$P(X = 5) = \frac{1470.08443 * 0.01356859825}{120}$

$P(X = 5) = \frac{19.9469850243}{120}$

$P(X = 5) = 0.1662248752$

$P(X = 5) = 0.166$ Approximated

6 0

2 years ago

what are two decimals with thousandths that would give 0.2 when rounded to the nearest tenth and 0.20 when rounded to the neares

egoroff_w [7]

Ur answer it 0.04 bc u would take the 0.2•0.20=0.04

3 0

3 years ago

A psychologist would like to examine the effect of fatigue on mental alertness. An attention test is prepared which requires sub

lianna [129]

Answer:

The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that fatigue does not significantly increase errors on an attention task at 0.05 significance level

Step-by-step explanation:

We formulate null and alternate hypotheses are

H0 : u1 < u2 against Ha: u1 ≥ u 2

Where u1 is the group tested after they were awake for 24 hours.

The Significance level alpha is chosen to be ∝ = 0.05

The critical region t ≥ t (0.05, 13) = 1.77

Degrees of freedom is calculated df = υ= n1+n2- 2= 5+10-2= 13

Here the difference between the sample means is x`1- x`2= 35-24= 11

The pooled estimate for the common variance σ² is

Sp² = 1/n1+n2 -2 [ ∑ (x1i - x1`)² + ∑ (x2j - x`2)²]

= 1/13 [ 120²+360²]

Sp = 105.25

The test statistic is

t = (x`1- x` ) /. Sp √1/n1 + 1/n2

t= 11/ 105.25 √1/5+ 1/10

t= 11/57.65

t= 0.1908

The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that fatigue does not significantly increase errors on an attention task at 0.05 significance level

5 0

2 years ago

Find the domain restriction for the function f(x)=(2x-5)^4. Use an inequality to state the domain.

Natasha_Volkova [10]

(2x-5x)^4+2x-16 f is 16

8 0

2 years ago

**URGENT**please help and show work!

Natasha2012 [34]

Answer:

mat hhway u gives all the answers

Step-by-step explanation: