Usman458
12.08.2020 •
Mathematics
normal population has a mean of 63 and a standard deviation of 13. You select a random sample of 25. Compute the probability that the sample mean is: (Round your z values to 2 decimal places and final answers to 4 decimal places): Greater than 65.
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Ответ:
0.2207
Step-by-step explanation:
Here, we want to find the probability that the sample mean is greater than 25.
What we use here is the z-scores statistic
Mathematically;
z-score = (x-mean)/SD/√n
From the question;
x = 65, mean = 63, SD = 13 and n = 25
Plugging these values in the z-score equation, we have
Z-score = (65-63)/13/√25 = 2/13/5 = 0.77
So the probability we want to calculate is ;
P(z > 0.77)
This can be obtained from the standard normal distribution table
Thus;
P(z > 0.77) = 0.22065 which is 0.2207 to 4 d.p
Ответ:
A partir del problemas vamos a crear una ecuación donde X es una docena de huevos y Y una libra de mantequilla
entonces: 4 docenas de huevos y 3 libras de mantequilla es igual a 14.100
en mi primera ecuación seria: 4X + 3Y = 14.100
ahora haremos la segunda ecuacion: tres docenas y 1 libra de mantequilla
en mi segunda ecuación seria: 3X + 1Y = 8.700 el plan es sumar las dos ecuaciones pero yo quiero que al sumarlas la variable "Y" desaparezca así que mi segunda ecuación la voy a multiplicar pór (-3) quedándome todo así:
primera ecuación: 4X + 3Y = 14.100
segunda ecuacion multiplicada por -3: -9X - 3Y = -26.1
al sumarlas obtengo como resultado: -5X = -12
resolviendo esa ecuacion obtengo que X=
X=2.4
luego reemplazas X en la primera ecuación: 4X + 3Y = 14.100
reemplazando seria: 4 x (2.4) + 3Y =14.100
9.6 + 3Y = 14.1000
3Y= 14.100 - 9.6
Y=
Y=1.5
entonces concluimos que la docena de huevos cuesta 2.4 $ y la libra de mantequilla 1.5 $
Step-by-step explanation: