"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?"
I think the answer is 16%. Could someone check this for me?
"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 think the answer for this one is 47.5%, so could someone also check this one?
There were answers to your previous questions. Perhaps you could review the answers, and if you need further help, and state where you need clarifications with reference to the previous answers.
http://www.jiskha.com/display.cgi?id=1337853463
and perhaps:
http://www.jiskha.com/display.cgi?id=1337912757
To find the probability that a randomly selected high jumper has a personal best between 229 and 248 cm, we need to calculate the area under the normal distribution curve between these two values.
First, we need to standardize the values by converting them to z-scores. The formula to calculate the z-score is (x - mean) / standard deviation.
For 229 cm:
z1 = (229 - 229) / 19 = 0
For 248 cm:
z2 = (248 - 229) / 19 = 19 / 19 = 1
Next, we need to find the area under the standard normal distribution curve between 0 and 1. To do this, we can use a standard normal distribution table or a calculator.
Using a standard normal distribution table, the area between 0 and 1 is approximately 0.3413.
Since the area under the normal distribution curve is symmetrical, we can double this value to find the area between -1 and 1, which encompasses the range from 229 to 248 cm.
0.3413 * 2 = 0.6826
Therefore, the probability that a randomly selected high jumper has a personal best between 229 and 248 cm is approximately 0.6826 or 68.26%, not 16%.
For the second question, the probability of finding between 102 and 117.5 employees who use the internet while at work can be calculated using the binomial distribution formula.
The binomial distribution formula is P(x) = (nCx) * (p^x) * (q^(n-x)), where n is the number of trials (respondents), x is the number of successes (employees who use the internet), p is the probability of success (47% or 0.47), q is the probability of failure (1 - p), and nCx represents the combination of n items taken x at a time.
In this case, n = 250, p = 0.47, and we want to find the probability of having between 102 and 117.5 employees using the internet (x = 102 to 117).
To find the probability, we need to calculate the cumulative probabilities for x = 102 to 117. We can use a binomial probability calculator or a statistical software to calculate this.
Using a binomial probability calculator, the probability of finding between 102 and 117 employees who use the internet while at work is approximately 47.5%, so your answer is correct.