Assume that women's heights are normally distributed with a mean given by mu equals 63.2 in, and a standard deviation given by
sigma equals 2.1 in. complete parts a and
b.
a. if 1 woman is randomly selected, find the probability that her height is between 62.4 in and 63.4 in. the probability is approximately . 1863. (round to four decimal places as needed.)
b. if 7 women are randomly selected, find the probability that they have a mean height between 62.4 in and 63.4 in
b.
a. if 1 woman is randomly selected, find the probability that her height is between 62.4 in and 63.4 in. the probability is approximately . 1863. (round to four decimal places as needed.)
b. if 7 women are randomly selected, find the probability that they have a mean height between 62.4 in and 63.4 in
1 answer:
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Answer:
A) See the picture
B) 14
C) 45%
Step-by-step explanation:
A) To create a histogram like the one on the picture you can use an online tool like socscistatistics where the number of classes is customizable
B) Because the question B and C have to be responded using a frequency table with 8 classes the answer is 14; the method of using cumulative frequency tables should only be considered as a way of estimation, that is because you obtain values that depend on your choice of class intervals. The way to get a better answer would be to use all the scores in the distribution
Pc1 = 100*(4/40) = 10
Pc2 = 100*(4/40) = 10
Pc3 = 100*(3/40) = 7.5
Pc4 = 100*(11/40) = 27.5
Pc5 = 100*(5/40) = 12.5
Pc6 = 100*(4/40) = 10
Pc7 = 100*(7/40) = 17.5
Pc8 = 100*(2/40) = 5
Pc8 + Pc7 + Pc6 + Pc5 + Pc4 + Pc3 + Pc2 = 90%
Therefore, From class 8 to class 2 is the top 90% of the applicants and the minimum score is 14.
C) Scores equal to or greater than 20 are from class 8 to class 5
Pc8 + Pc7 + Pc6 + Pc5 = 45%
