Last year, the personal best high jumps of track athletes in a nearby state were normally distributed with a mean of 229 cm and a standard deviation of 19 cm. What is the probability that a randomly selected high jumper has a person best between 229 and 248 cm?
According to a poll conducted in a company, 47% of the employees regularly use the internet while at work. You conduct a similar survey at your mother's office where there are 250 respondents. What is the probability you find between 102 and 117.5 employees who use the internet while at work?
I have no idea how to solve either one of these problems, could someone please show me the steps?
First Problem
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.
To solve both of these problems, we will use the concept of a normal distribution. A normal distribution, also known as a bell curve, is a continuous probability distribution that is symmetric and follows a specific mathematical formula. In mathematics, this distribution is often denoted by the symbol "μ" (mean) and "σ" (standard deviation).
To find the probability that a randomly selected high jumper has a personal best between two specific values, we can use the standard normal distribution table or the Z-score formula.
1. High Jumpers' Personal Best:
Given: Mean (μ) = 229 cm, Standard Deviation (σ) = 19 cm
We are looking for the probability between 229 cm and 248 cm.
Step 1: Calculate the Z-scores
Z1 = (x1 - μ) / σ
Z2 = (x2 - μ) / σ
Here, x1 = 229 cm, x2 = 248 cm
Z1 = (229 - 229) / 19 = 0
Z2 = (248 - 229) / 19 ≈ 1
Step 2: Use the Z-table or calculator
Look up the probability values for Z1 and Z2 from the standard normal distribution table or use a calculator.
P(Z < 0) = 0.5000
P(Z < 1) ≈ 0.8413
Step 3: Calculate the probability
The probability between 229 cm and 248 cm is given by:
P(229 cm < X < 248 cm) = P(Z < Z2) - P(Z < Z1) = 0.8413 - 0.5000 ≈ 0.3413
So, there is approximately a 34.13% probability that a randomly selected high jumper has a personal best between 229 cm and 248 cm.
2. Internet Usage at the Office:
Given: Percentage of internet users (p) = 47%, Number of respondents (n) = 250
We are looking for the probability between 102 and 117.5 employees who use the internet at work.
Step 1: Calculate the mean (μ) and standard deviation (σ) of the binomial distribution.
For a binomial distribution, the mean (μ) is calculated as μ = n * p
And the standard deviation (σ) is calculated as σ = √(n * p * (1 - p))
Here, p = 47%, n = 250
μ = 250 * 0.47 = 117.5
σ = √(250 * 0.47 * (1 - 0.47)) ≈ 8.574
Step 2: Calculate the Z-scores
Z1 = (x1 - μ) / σ
Z2 = (x2 - μ) / σ
Here, x1 = 102, x2 = 117.5
Z1 = (102 - 117.5) / 8.574 ≈ -1.814
Z2 = (117.5 - 117.5) / 8.574 = 0
Step 3: Use the Z-table or calculator
Look up the probability values for Z1 and Z2 from the standard normal distribution table or use a calculator.
P(Z < -1.814) ≈ 0.0351
P(Z < 0) = 0.5000
Step 4: Calculate the probability
The probability between 102 and 117.5 employees who use the internet at work is given by:
P(102 < X < 117.5) = P(Z < Z2) - P(Z < Z1) = 0.5000 - 0.0351 ≈ 0.4649
So, there is approximately a 46.49% probability that between 102 and 117.5 employees at your mother's office use the internet while at work.