A study on the latest fad diet claimed that the amounts of weight lost by all people on this diet had a mean of 24.6 pounds and a standard deviation of 8.0 pounds.

Step 2 of 2 : If a sampling distribution is created using samples of the amounts of weight lost by 51 people on this diet, what would be the standard deviation of the sampling distribution of sample means? Round to two decimal places, if necessary.

Answer:

the standard deviation of the sampling distribution of sample means is 1.12

Step-by-step explanation:

Given the data in the question,

population mean; μ = 24.6 pounds

Population standard deviation; σ = 8.0 pounds

sample size; n = 51

Now determine the standard deviation of the sampling distribution of sample means.

standard deviation of the sampling distribution of sample means is simply

⇒ population standard deviation / √sample size

= 8.0 / √51

= 8.0 / 7.141428

= 1.120224 ≈ 1.12

Therefore, the standard deviation of the sampling distribution of sample means is 1.12

5 0

3 years ago

The local water slides have 100 employees, of which 30% are temporary. How many temporary employees are there?

Ugo [173]

Answer:

There are 30 temporary employees

Step-by-step explanation:

To find x% of quantity A, multiply x% by A after change x% to the normal number by divide it by 100 ⇒ (x ÷ 100) × A

Let us use this rule to solve the question

∵ The local water slides have 100 employees

∵ 30% of them are temporary

∴ The number of temporary employees = 30% × 100

∵ 30% = 30 ÷ 100 = 0.30

∴ The number of temporary employees = 0.30 × 100

∴ The number of temporary employees = 30

∴ There are 30 temporary employees

3 0

3 years ago

What are the solutions to the equation (2x-5)(3x-1)=0?

lutik1710 [3]

(2x-5)(3x-1)=0

2x-5 = 0 or 3x-1= 0
2x = 5 3x = 1
x = 5/2 x = 1/3

Answer: x=5/2; x=1/3

6 0

3 years ago

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