Answer: 0.5056
Step-by-step explanation:
Let x be the random variable that represents the readings on scientific thermometers .
Given : The men's weights are normally distributed,
Population mean : [tex]\mu=172\ lb[/tex]
Standard deviation : [tex]\sigma= 29\ lb[/tex]
Sample size : n=4
Z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
Now, the z-value corresponding to 160 : [tex]z=\dfrac{160-172}{\dfrac{29}{\sqrt{4}}}\approx-0.83[/tex]
z-value corresponding to 180 : [tex]z=\dfrac{180-172}{\dfrac{29}{\sqrt{4}}}\approx0.55[/tex]
By using standard normal distribution table for z-values,
P-value = [tex]P(160<x<180)=P(-0.83<z<0.55)=P(z<0.55)-P(z<-0.83)[/tex]
[tex]= 0.7088403-0.2032694=0.5055709\approx0.5056\text{ (Rounded to four decimal places)}[/tex]
Hence, the probability that they have a mean weight between 160 lb and 180 lb = 0.5056
Hence, the probability of the reading greater than -1.05 in degrees Celsius.= 0.8531
There is a probability of 26.94% that they have a mean weight between 160 lb and 180 lb
Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
z = (raw score - mean)/standard deviation
Given that:
Mean = 172 lb, standard deviation = 29 lb, For 160:
z = (160 - 172) / 29 = -0.41
For 180:
z = (180 - 172) / 29 = 0.28
P(160 < x < 180) = P(z < 0.28) - P(z < -0.41) = 0.6103 - 0.3409 = 0.2694
There is a probability of 26.94% that they have a mean weight between 160 lb and 180 lb
Find out more on z score at: https://brainly.com/question/25638875
