In a random sample of 50 undergraduate students at a college, it was found that 44 students regularly access social networking websites from their college library. what is the margin of error for the true proportion of all undergraduates who access social networking sites from their college library? hurry

Margin of Error = 9.01%

Explanation:

Given

Sample size, n = 50

Number of regular access = 44

The level of confidence is not stated.

I'll assume a 95% level of confidence, so z* = 1.96

To calculate Margin of error,

First we need to calculate the proportion of those that have regular access and those that don't, as probabilities.

Let P = Probability of those with regular access.

P = 44/50 = 0.88

Let Q = probability of those that don't have regular access

P + Q = 1 --- make Q the subject of formula

Q = 1 - P

Q = 1 - 0.88

Q = 0.12

First, we calculate the standard error

Standard Erro is calculated using the following formula.

A.E = √(PQ/n)

By substituton

S.E = √(0.88*0.12/50)

S.E = 0.045956501172304

Finally,. margin of error is calculated by S.E * z*

= 0.045956501172304 * 1.96

Margin of Error = 0.090074742297716

Margin of Error = 9.01% --- Approximated

the inequality is:

250c+180g≥950

step-by-step explanation:

let c denotes the number of days in a week   miyoko worked as a cryptographer.

and g denotes the number of days in a week she works as a geologist.

now we are given that she earns $250 for every day she works as a cryptographer,and earns $180 for every day she works as a geologist.

now miyoko's goal is to earn more than $950 this week.

hence, the inequality could be represented as:

250c+180g≥950

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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