Find the probability that an observation from a normally distributed random vari
ID: 3173955 • Letter: F
Question
Find the probability that an observation from a normally distributed random variable will fall within 1.2 standard deviations of its mean. Find the probability that an observation from a normally distributed random variable will fall within 2.8 deviations of its mean. Research has demonstrated that milk consumption for American households is normally distributed with the average family consuming 132 ounces per week with a standard deviation of 9 ounces. What percentage of household consume more than a 160 ounces of milk per week? The average number of medication errors per hospital per day is normally distributed with a mean of 4 per day with a standard deviation of 1.2. What percentage of hospitals have less than 3 errors per day?Explanation / Answer
Q.1 Normal distribution so that value will fall within 1.2 times the standerd deviation so z value of it = +- 1.2
so relative probabibilities first for Z = -1.2; P( x < mean - 1.2 std. dev) = 0.1151
so Z = + 1.2; P ( x > mean + 1.2 std. dev) = 0.8849
so total probability = ( 0.5 - 0.1151) + ( 0.8849 - 0.5) = 0.7698 = 76.98 %
Q.2 Here Z values are +- 2.8
so relative probabibilities first for Z = -2.8; P( x < mean - 2.8 std. dev) = 0.0026
so Z = + 2.8; P ( x > mean + 2.8 std. dev) = 0.9974
so total probability = ( 0.5 - 0.0026) + ( 0.9974 - 0.5) = 0.9948 = 99.48 %
Q.3 Mean = 132 ounces per week ; standerd deviation = 9 ounces
P ( x > 160; 132 ; 9) so calculate the z- value = (160 -132)/9 = 3.11
P ( x < 160; 132 ; 9) for the given z value = 0.9991
so P ( x > 160; 132 ; 9) = 1 - 0.9991 = 0.0009
Q.4 Average of medication errors = 4 per day
standerd deviation = 1.2
we have to calculate error below 3 per day
so P( x <3; 4; 1.2) ; z value = (3-4) /1.2 = - 0.833
P (x <3; 4;1.2) from z -table = 0.2023 = 20.23 %
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