As we don't know the standard deviation of the population, we can estimate the standard error of the mean from the standard deviation of the sample as:

$\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}$

The sample is [30mins, 40 mins, 60 mins, 80 mins, 20 mins, 85 mins]. The size of the sample is n=6.

The mean of the sample is:

$\bar{x}=\frac{1}n} \sum x_i =\frac{30+40+60+80+20+85}{6}=52.5$

The standard deviation of the sample is calculated as:

$s=\sqrt{\frac{1}{n-1}\sum (x_i-\bar x)^2} \\\\ s=\sqrt{\frac{1}{5}\cdot ((30-52.5)^2+(40-52.5)^2+(60-52.5)^2+(80-52.5)^2+(20-52.5)^2+(85-52.5)^2}\\\\s=\sqrt{\frac{1}{5} *3587.5}=\sqrt{717.5}=26.8$

Then, we can calculate the standard error of the mean as:

$\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}=\frac{26.8}{6}= 4.5$

6 0

2 years ago

PLEASE HELPP BRANLIEST AND MEDAL PLSSS

Fudgin [204]

56. and 1.6^4

.....................

8 0

2 years ago

The sum of two numbers is less than 2. If we subtract the second number from the first, the difference is greater than 1. What a

Anna007 [38]

Answer:

the value of these two numbers are 1/2 and 3/2

Step-by-step explanation:

Given that:

x+y < 2 ---- (1)

y - x > 1 ---- (2)

From equation (2), let y > 1 + x then substitute it into equation (1)

x + 1 + x < 2

2x + 1 < 2

2x < 2 - 1

2x < 1

x < 1/2

From equation (2), replace the value of x to be 1/2

y - 1/2 > 1

y > 1 + 1/2

y > 3/2

5 0

2 years ago

Kylie draws a triangle on a coordinate grid. She then performs the following

zmey [24]

B PLEASE GIVE BRAINLIEST

7 0

2 years ago