Answer:
1/8
Step-by-step explanation:
Answer:
Step-by-step explanation:
Hello!
To study the threshold of hering the researcher took a random sample of 80 male college freshmen.
The students underwent an audiometry test where a tome was played and they had to press a button when they detected it. The researcher recorded the lowest stimulus level at which the tone was detected obtaining a sample mean of X[bar]= 22.2 dB and a standard deviation of S= 2.1 dB
To estimate the population mean, since we don't have information about the variable distribution but the sample size is greater than 30, you can use the approximation of the standard normal distribution:
X[bar] ± 
Where the semiamplitude or margin of error of the interval is:
d= 
Using a 95% level 
d= 1.965 * 
d= 0.46
The point estimate of the population mean of the threshold of hearing for male college freshmen is X[bar]= 22.2 db
And the estimation using a 95%CI is [21.74;22.66]
I hope this helps!
Answer:
stop rushing brat
Step-by-step explanation:
Answer: 
Step-by-step explanation:
The area of a rectangle can be calculated with the formula:

l: the length of the rectangle.
w: the width of the rectangle.
The area of the remaning wall after the mural has been painted, will be the difference of the area of the wall and the area of the mural.
Knowing that the dimensions of the wall are
by
, its area is:

As they are planning that the dimensions of the mural be
by
, its area is:

Then the area of the remaining wall after the mural has been painted is:

Answer:
Step-by-step explanation:
Hello!
The variable of interest is:
X: number of daily text messages a high school girl sends.
This variable has a population standard deviation of 20 text messages.
A sample of 50 high school girls is taken.
The is no information about the variable distribution, but since the sample is large enough, n ≥ 30, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:
X[bar]≈N(μ;δ²/n)
This way you can use an approximation of the standard normal to calculate the asked probabilities of the sample mean of daily text messages of high school girls:
Z=(X[bar]-μ)/(δ/√n)≈ N(0;1)
a.
P(X[bar]<95) = P(Z<(95-100)/(20/√50))= P(Z<-1.77)= 0.03836
b.
P(95≤X[bar]≤105)= P(X[bar]≤105)-P(X[bar]≤95)
P(Z≤(105-100)/(20/√50))-P(Z≤(95-100)/(20/√50))= P(Z≤1.77)-P(Z≤-1.77)= 0.96164-0.03836= 0.92328
I hope you have a SUPER day!