Statistics Assignment: Normal Distribution and Sample Analysis

This assignment solution addresses several statistical problems related to normal and sampling distributions. The first set of problems concerns the diameter of tennis balls, utilizing the normal distribution to calculate probabilities associated with sample means and determining values symmetrically distributed around the population mean. The central limit theorem is applied to analyze the sampling distribution of the mean. The second set of problems focuses on the time spent on a social media platform, Tumblr. Probabilities are calculated for the sample mean falling within specified ranges, considering different sample sizes. The solutions involve calculating Z-scores and using the standard normal distribution to find probabilities. Overall, the assignment demonstrates the application of statistical concepts to real-world scenarios.

Running head: STATISTICS
STATISTICS
Name of the Student
Name of the University
Author Note

Q 7.5:
1.
Given mean diameter of tennis balls μ = 2.63 inches
Standard deviation σ = 0.03 inch.
Also, diameter ~ N(2.63,0.03)
Now, a random sample of 9 tennis balls are selected.
Now, by the central limit theorem the sampling distribution of mean has the mean of μ = 2.63
inches and standard deviation of σ/sqrt(9) = 0.03/3 = 0.01. As the sample size increases the
sample distribution of mean reaches to normal distribution.
2.
Let, the sample mean = Xbar
Hence, P(X<2.61) = P(Z<(2.61-2.63)/0.01) = P(Z<-2) = 0.02275 or 2.275%.
3.
P(2.62<X<2.64) = P(X<2.64) – P(X<2.62) = P(Z<(2.64-2.63)/0.01) – P(X<(2.62-2.63)/0.01)
= 0.6827 or 68.27%.
4.
Probability between two values symmetrically distributed around population mean is 60% or
the probability over the larger value is (100-60)/2 = 20% as normal distribution is symmetric.
Let, the larger value is Z2.
Hence, P(Z>Z2) = 20% or 0.2 = 1- P(Z<Z2) = 0.2
 P(Z<Z2) = 0.8

Now, from standard normal distribution the value of Z2 under which 80% of area lies is Z2 =
0.8416.
Hence, as normal distribution is symmetric about its mean hence the lower Z value is -
0.8416.
Hence, X1 = 2.63 -0.8416*0.01 = 2.6216
Now, X2 = 2.63 + 0.8416*0.01 = 2.6384.
Hence, the two values are 2.6216 and 2.6384 symmetrically distributed over population mean
of 2.63 inside which probability is 60%.
Q.7.7
Time spent ~ N(28,5)
n = 25.
Hence, sampling distribution of mean ~ N(28,5/sqrt(25)) = N(28,1)
27.
Let, sample mean is Xbar.
P(27.5<Xbar<28.5) = P(Z<(28.5-28)) – P(Z<(27.5-28)) = P(Z<0.5) – P(Z<-0.5)
= 0.3829 or 38.29%.
28.
P(27<Xbar<29) = P(Z<(29-28)) – P(Z<(27-28)) = P(Z<1) – P(Z<-1) = 0.6827 = 68.27%.
29.
Now, n = 100.

Hence, sampling distribution of mean has the mean of 28 and standard deviation =
5/sqrt(100) = 5/10 = 0.5.
Hence, P(27.5<Xbar<28.5) = P(Z<(28.5-28)/0.5) – P(Z<(27.5-28)/0.5) = P(Z<1) – P(Z<-1)
= 0.6827 or 68.27%.
Q 7.14
1.
Given, mean proportion μp = 74% or 0.74.
Hence, p = 0.74. Sample size = 100.
np = 100*0.74 = 74 > 10.
Hence, as np>10 the sampling distribution of sample proportion can be assumed to be
approximately normally distributed with mean μp = 0.74 and standard deviation σp =
√ p ( 1− p )
n
=
√ 0.74 ( 1−0.74 )
100 = 0.0439
Hence, the probability that fewer than 78% say that they will be able to work flexibly and still
be on track for promotion is important is
P(Xp<0.78) = P(Z<(0.78-0.74)/0.0439) = P(Z<0.9111) = 0.8189 or 81.89%.
2.
P(0.70<Xp<0.78) = P(Z<(0.78-0.74)/0.0439) – P(Z<(0.70-0.74)/0.0439) = P(Z<0.9111) –
P(Z<-0.9111) = 0.6378 or 63.78%.

3.
P(Xp>0.76) = 1- P(Xp<0.76) = 1 – P(Z< (0.76-0.74)/0.0439) = 1 – P(Z< 0.4556) = 0.3243 or
32.43%.