| Date | May 2018 | Marks available | 5 | Reference code | 18M.2.sl.TZ2.10 |
| Level | SL only | Paper | 2 | Time zone | TZ2 |
| Command term | Find | Question number | 10 | Adapted from | N/A |
Question
The mass M of apples in grams is normally distributed with mean μ. The following table shows probabilities for values of M.
The apples are packed in bags of ten.
Any apples with a mass less than 95 g are classified as small.
Write down the value of k.
Show that μ = 106.
Find P(M < 95) .
Find the probability that a bag of apples selected at random contains at most one small apple.
Find the expected number of bags in this crate that contain at most one small apple.
Find the probability that at least 48 bags in this crate contain at most one small apple.
Markscheme
evidence of using \(\sum {{p_i}} = 1\) (M1)
eg k + 0.98 + 0.01 = 1
k = 0.01 A1 N2
[2 marks]
recognizing that 93 and 119 are symmetrical about μ (M1)
eg μ is midpoint of 93 and 119
correct working to find μ A1
\(\frac{{119 + 93}}{2}\)
μ = 106 AG N0
[2 marks]
finding standardized value for 93 or 119 (A1)
eg z = −2.32634, z = 2.32634
correct substitution using their z value (A1)
eg \(\frac{{93 - 106}}{\sigma } = - 2.32634,\,\,\frac{{119 - 106}}{{2.32634}} = \sigma \)
σ = 5.58815 (A1)
0.024508
P(X < 95) = 0.0245 A2 N3
[5 marks]
evidence of recognizing binomial (M1)
eg 10, ananaCpqn−=××and 0.024B(5,,)pnp=
valid approach (M1)
eg P(1),P(0)P(1)XXX≤=+=
0.976285
0.976 A1 N2
[3 marks]
recognizing new binomial probability (M1)
eg B(50, 0.976)
correct substitution (A1)
eg E(X) = 50 (0.976285)
48.81425
48.8 A1 N2
[3 marks]
valid approach (M1)
eg P(X ≥ 48), 1 − P(X ≤ 47)
0.884688
0.885 A1 N2
[2 marks]
