z = (x - mean_x)/standard_deviation
→ z = (51 - 57)/3.5 = -1.71
Being negative, it shows it is less than the mean; only the absolute value needs to be considered
Looking up z = 1.71 in normal tables gives a value of 0.4564 which is the probability that the value lies between 51 and 57, so the probability of it being less than 51 is 0.5 - 0.4564 = 0.0436 = 4.36 %
The probability of getting less than 51 mpg is approx 4.36 %
Approx 4.3237%
(X-61)/10.2=.878 X = 69.95 approximately 70
There are approximately 16.4% of students who score below 66 on the exam.
68.2%
A particular fruit's weights are normally distributed, with a mean of 760 grams and a standard deviation of 15 grams. If you pick one fruit at random, what is the probability that it will weigh between 722 grams and 746 grams-----A particular fruit's weights are normally distributed, with a mean of 567 grams and a standard deviation of 25 grams.
True.
A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounce. Find the proportion of all jars packaged by this process that have weights that fall below 10.875 ounces.
The mean and standard deviation. If the data really are normally distributed, all other statistics are redundant.
BMI varies from person to person and one measure of this variability is the standard deviation. Assuming the BMI is approximately normally distributed, only around 0.135% of the people will have results that are -3 sd or lower.
It the upper bound of the range is U, thenZL = (10.25 - 10.75)/0.8 = -0.625 and ZU = (U - 10.75)/0.8 Then the relevant proportion can be determined as the integral of the standard normal distribution between ZL and ZU.
Yes, to approximately standard normal.If the random variable X is approximately normal with mean m and standard deviation s, then(X - m)/sis approximately standard normal.
94.99% of an 8oz. cup
The Miller Analogies Test scores have a mean of 400 and a standard deviation of 25, and are approximately normally distributed.z = ( 351.5 - 400 ) / 25 = -1.94That's about the 2.6 percentile.(Used wolframalpha.com with input Pr [x < -1.94] with x normally distributed with mean 0 and standard deviation 1.)
(X-61)/10.2=.878 X = 69.95 approximately 70
There are approximately 16.4% of students who score below 66 on the exam.
68.2%
A particular fruit's weights are normally distributed, with a mean of 760 grams and a standard deviation of 15 grams. If you pick one fruit at random, what is the probability that it will weigh between 722 grams and 746 grams-----A particular fruit's weights are normally distributed, with a mean of 567 grams and a standard deviation of 25 grams.
d. All the above.