1. The number of pizzas consumed per month by university students is normally distributed with a mean of 12 and a standard deviation of 4. What is the probability that a randomly selected sample of 25 students will consume on average more than 14 pizzas per month?

0.00621

Step-by-step explanation:

To solve for this question: we would be making use of the z score formula.

z = (x - μ)/σ

where

x is the raw score

μ is the population mean

σ is the population standard deviation

Step 1

Find the Standard Error

From the above question, we are given number of samples, hence, the standard deviation we would use for our z score is Standard error.

Standard error = Standard deviation/√n

Where n = number of samples = 25

Standard deviation = 4

Standard error = 4/√25

= 4/5

= 0.8

Step 2

Calculate the z score

Since we are given number of samples in the question,

z score = z = (x - μ)/σ

where

x is the raw score = 14

μ is the population mean = 12

σ is the standard error = 0.8

z score = 14 - 12/0.8

z score = 2.5

Step 3

We find the Probability.

The probability that a randomly selected sample of 25 students will consume on average more than 14 pizzas per month is calculated as:

We find the Probability of the Z score 2.5 using a Z table.

P(z = 2.5) = P(x ≤ 14) = 0.99379

P(x > 14) = 1 - P(x<14)

= 1 - 0.99379

= 0.0062097

Approximately ≈ 0.00621

Therefore, the probability that a randomly selected sample of 25 students will consume on average more than 14 pizzas per month is

0.00621

is should be y = 5x

step-by-step explanation:

correct me if im wrong

u = -(6/7)w

step-by-step explanation:

place all values with 'u' on one side of the equation.

4u + 8w = - 3u + 2w

4u + 3u = 2w - 8w

further simplify the equation by solving the addition and subtraction.

7u = - 6w

make 'u' the independent variable by placing its co-efficient on the right-hand side.

u = - (6/7)w