The mean output of a certain type of amplifier is 352352 watts with a variance of 121121. If 5757 amplifiers are sampled, what is the probability that the mean of the sample would differ from the population mean by more than 2.32.3 watts

88.56%

Step-by-step explanation:

given:

Mean M = 352

Variance V = 121

sample size n = 57

. To find p(x <2..32 )

for that we use the formula of normal distribution and the normal distribution table

First: The normal distribution formula of Z score is :

z = (x - M ) / (SD / sqrt(n))

. SD = sqrt(variance)

SD = sqrt(121)

SD = 11

On plugging the values we get

z = (352-2.32) / (11/sqrt(57))

z = 1.58

Second, we refer to the Z table for probability value

P(z < 1.58) can be obtained from the Z table

=0.8858.

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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