Question

Please answer all parts as I know the answers but need the work to go with them. Thus, I believe the above answers are correct. Thank you!

Answer 1

As the sample size n is less than 30 normal distribution is used.

The values of x are converted into z by using the formula z= x-u/ s and then the z values are found out from the table.

The limits are found by using the formula x±σz or x±sz where s= σ

As the sample size is 10 which is less than 30 the normal distribution is used.

The probability of x< 2.59 is 0.3446

The probability 2.60<X <2.63 is 0.9484

So lower and upper limits are 2.607 and 2.612

Part A

As the sample size is 10 which is less than 30 the normal distribution is used.

Part B

For given value of x= 2.59 z is obtained =0.4

x= 2.59

z= x-u/ s

z= 2.59-2.61/0.05

z= -0.02/0.05

z=- 0.4

P (X<2.59) = P(-0.4 <Z<0) = 0.5 -0.1554= 0.3446

The probability of x< 2.59 is 0.3446

Part C

For two given values of x= 2.60 and 2.63 z is obtained as =0.2 and 0.4

x1= 2.60

z1= x-u/ s

z= 2.60-2.61/0.05

z= -0.01/0.05

z=- 0.2

x2= 2.63

z2= x-u/ s

z= 2.63-2.61/0.05

z= 0.02/0.05

z= 0.4

P (2.60<X<2.63) = P(-0.2 <Z<0.4)

= P(-0.2 <Z<0)+ P(0 <Z<0.4)

=0.793 + 0.1554= 0.9484

The probability 2.60<X <2.63 is 0.9484

Part D"

p= 0.57

From the table z= 0.045

z= x-u/ s

zs= x-u

zs+u = x

x1= 0.045*0.05 +2.61= 2.61225

x2= 2.61- 0.00225= 2.60775

So lower and upper limits are 2.607 and 2.612

For further understanding of probability calculation click

brainly.com/question/25638875

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