Question

The higher the bowling score the better. The lower the golf score the better. Assume both are normally distributed. a. Suppose we have a sample of the Santa Ana Strikers' bowling scores. Q1 = 125 and Q3 = 156. Would it be usual or unusual to have a score of 200?b. Suppose the mean bowling score is 155 with a standard deviation of 16 points. What is the probability that in a sample of 40 bowling scores, the mean will be smaller than 150?c. Suppose the mean golf score is 77 with a standard deviation of 3 strokes We will give a trophy for the best 5% of scores. What score must you get to receive a trophy? d. Suppose the mean golf score is 77 with a standard deviation of 3 strokes. Would a golf score of 70 be ordinary, a mild outlier, or an extreme outlier?

Answer 1

Answer:

Explained below.

Step-by-step explanation:

(a)

The first and third quartiles of bowling scores are as follows:

Q₁ = 125 and Q₃ = 156

Then the inter quartile range will be:

IQR = Q₁ - Q₃

      = 156 - 125

      = 31

Any value lying outside the range (Q₁ - 1.5×IQR, Q₃ + 1.5×IQR) are considered as unusual.

The range is:

(Q₁ - 1.5×IQR, Q₃ + 1.5×IQR) = (125 - 1.5×31, 156 + 1.5×31)

                                               = (78.5, 202.5)

The bowling score of 200 lies in this range.

Thus, the bowling score of 200 is usual.

(b)

Compute the probability that the mean bowling score will be smaller than 150 as follows:

$P(\bar X$

                  $=P(Z$

Thus, the probability that in a sample of 40 bowling scores, the mean will be smaller than 150 is 0.024.

(c)

It is provided that, the lower the golf score the better.  

So, the best 5% of scores would be the bottom 5%.

That is, P (X > x) = 0.05.

⇒ P (Z > z) = 0.05

⇒ P (Z < z) = 0.95

⇒ z = 1.645

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\\\1.645=\frac{x-77}{3}\\\\x=77+(3\times 1.645)\\\\x=81.935\\\\x\approx 82$

Thus, the score is 82.

(d)

A z-scores outside the range (-2, +2) are considered as mild outlier and the z-scores outside the range (-3, +3) are considered as extreme outlier.

Compute the z-score for the golf score of 70 as follows:

$z=\frac{x-\mu}{\sigma}$

  $=\farc{70-77}{3}\\\\=\frac{-7}{3}\\\\=-2.33$

As the z-score for the golf score of 70 is less than -2, it is considered as a mild outlier.