Answer: 0.9726
Step-by-step explanation:
Given : Females have pulse rates that are normally distributed with a mean of 74.0 beats per minute and a standard deviation of 12.5 beats per minutes.
i.e. and
Let x is a random variable to represent the pulse rates.
Formula :
For n= 16 , the probability that her pulse rate is less than 80 beats per minute will be :-
Hence, the required probability = 0.9726
The probability that a randomly selected female's pulse rate is less than 80 beats per minute, given a mean pulse rate of 74.0 and standard deviation of 12.5 beats per minute, is approximately 0.6844, or 68.44%.
This question pertains to the topic of normal distribution in statistics. We know that the average or mean pulse rate for females is 74.0 beats per minute, with a standard deviation of 12.5 beats per minute. We also know that the pulse rate we want to find the probability for is less than 80 beats per minute.
In these situations, we use the formula for the z-score, which is Z = (X - μ) / σ, where X is the value, we're interested in, μ is the mean, and σ is the standard deviation.
Using this formula, we find Z = (80 - 74) / 12.5 = 0.48. After finding the z-score, we can look at the standard normal distribution table to get the probability. The value for Z = 0.48 on the Z table is approximately 0.6844. Therefore, the probability that a randomly selected female's pulse rate is less than 80 beats per minute is approximately 0.6844, or 68.44%.
#SPJ6
Answer:
parllelogram
Step-by-step explanation:
............
h(x)=-49x-125
Half of the students like football.
The students would prefer to play sports over going to school.
None of the students like tennis.
Answer:
C. The students would prefer to play sports over going to school
Step-by-step explanation:
Hope this helps :)
can u brainlist
Answer:
Option E, two-proportion z test should be used to determine whether these data provide sufficient evidence to reject the hypothesis that the proportion of shoppers at the suburban mall who had been to a movie in the past month is the same as the proportion of shoppers in the large downtown shopping area who had been to a movie in the past month
Step-by-step explanation:
The complete question is
In a random sample of 60 shoppers chosen from the shoppers at a large suburban mall, 36 indicated that they had been to a movie in the past
month. In an independent random sample of 50 shoppers chosen from the shoppers in a large downtown shopping area, 31 indicated that
they had been to a movie in the past month. What significance test should be used to determine whether these data provide sufficient
evidence to reject the hypothesis that the proportion of shoppers at the suburban mall who had been to a movie in the past month is the same
as the proportion of shoppers in a large downtown shopping area who had been to a movie in the past month?
A one-proportion z interval B two-proportion z interval
B two-proportion z interval
C two-sample t test D one-proportion z test
D one-proportion z test
E two-proportion z test
Solution
Two proportion z test is used to compare two proportions. In this test the null hypothesis is that the two proportions are equal and the alternate hypothesis is that the proportions are not the same. The random sample of populations serve as two proportions.
Hence, option E is the best choice of answer